Degree of a differential equation is the power to which the highest derivative is raised when it is expressed as polynomial of derivatives.
Given equation is
$$\eqalign{ & {\left( {\frac{{{d^3}y}}{{d{x^3}}}} \right)^{\frac{2}{3}}} - 3\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right) + 5\left( {\frac{{dy}}{{dx}}} \right) + 4 = 0 \cr & \Rightarrow {\left( {\frac{{{d^3}y}}{{d{x^3}}}} \right)^{\frac{2}{3}}} = 3\frac{{{d^2}y}}{{d{x^2}}} - 5\frac{{dy}}{{dx}} - 4 \cr} $$
Cube on both side, $${\left( {\frac{{{d^3}y}}{{d{x^3}}}} \right)^2} = {\left[ {3\frac{{{d^2}y}}{{d{x^2}}} - 5\frac{{dy}}{{dx}} - 4} \right]^3}$$
Hence, degree $$ = 2.$$

Divide the equation by $$y{\left( {\log \,y} \right)^2}$$
$$\eqalign{ & \frac{1}{{y{{\left( {\log \,y} \right)}^2}}}.\frac{{dy}}{{dx}} + \frac{1}{{\log \,y}}.\frac{1}{x} = \frac{1}{{{x^2}}} \cr & {\text{Put }}\frac{1}{{\log \,y}} = z \Rightarrow \frac{{ - 1}}{{y{{\left( {\log \,y} \right)}^2}}}.\frac{{dy}}{{dx}} = \frac{{dz}}{{dx}} \cr & {\text{Thus, we get, }} - \frac{{dz}}{{dx}} + \frac{1}{x}.z = \frac{1}{{{x^2}}},{\text{ linear in }}z \cr & \Rightarrow \frac{{dz}}{{dx}} + \left( { - \frac{1}{x}} \right)z = - \frac{1}{{{x^2}}} \cr & {\text{I}}{\text{.F}}{\text{.}} = {e^{ - f\frac{1}{x}dx}} = {e^{ - \log \,x}} = \frac{1}{x} \cr & \therefore \,{\text{The solution is, }}z\left( {\frac{1}{x}} \right) = \int {\frac{{ - 1}}{{{x^2}}}\left( {\frac{1}{x}} \right)dx + c} \cr & \Rightarrow \frac{1}{{\log \,y}}\left( {\frac{1}{x}} \right) = \frac{{ - {x^{ - 2}}}}{{ - 2}} + c \cr & \Rightarrow x = \log \,y\left( {c{x^2} + \frac{1}{2}} \right) \cr} $$

The equation of parabolas having vertex at $$\left( {0,\,0} \right)$$  & focus at $$\left( {a,\,0} \right),$$  where $$\left( {a > 0} \right)$$  is :
$$\eqalign{ & {y^2} = 4ax......\left( 1 \right) \cr & 2y.\frac{{dy}}{{dx}} = 4a\,\,\,\,\,\,\,\,\,\left[ {{\text{on differentiating}}} \right] \cr} $$
On putting the value of $$\left( {4a} \right)$$  in equation $$\left( 1 \right)$$ we get,
$$\eqalign{ & 2x.\frac{{dy}}{{dx}} - y = 0 \cr & {\text{in order }} = 1{\text{ & degree }} = 1 \cr} $$

$$\eqalign{ & {\text{Putting }}v = \frac{y}{x}{\text{ so that }}x\frac{{dv}}{{dx}} + v = \frac{{dy}}{{dx}} \cr & {\text{We have }}x\frac{{dv}}{{dx}} + v = v + \phi \left( {\frac{1}{v}} \right) \cr & \Rightarrow \frac{{dv}}{{\phi \left( {\frac{1}{v}} \right)}} = \frac{{dx}}{x}\,; \cr & \Rightarrow \log \left| {Cx} \right| = \int {\frac{{dv}}{{\phi \left( {\frac{1}{v}} \right)}}} \cr & {\text{But }}y = \frac{x}{{\log \left| {Cx} \right|}}{\text{ is the general solution,}} \cr & {\text{So, }}\frac{x}{y} = \frac{1}{v} = \log \left| {Cx} \right| = \int {\frac{{dv}}{{\phi \left( {\frac{1}{v}} \right)}}} \cr & \Rightarrow \phi \left( {\frac{1}{v}} \right) = - \frac{1}{{{v^2}}}\,\,\,\,\left( {{\text{differentiating w}}{\text{.r}}{\text{.t}}{\text{. }}v{\text{ both sides}}} \right) \cr & \Rightarrow \phi \left( {\frac{x}{y}} \right) = - \frac{{{y^2}}}{{{x^2}}} \cr} $$

$$\eqalign{ & \frac{{dy}}{{dx}} = \sqrt {\frac{{1 - {y^2}}}{{1 - {x^2}}}} \cr & \therefore \,\frac{{dy}}{{\sqrt {1 - {y^2}} }} = \frac{{dx}}{{\sqrt {1 - {x^2}} }} \cr & \Rightarrow \int {\frac{{dy}}{{\sqrt {1 - {y^2}} }}} = \int {\frac{{dx}}{{\sqrt {1 - {x^2}} }}} \cr & \Rightarrow {\sin ^{ - 1}}y = {\sin ^{ - 1}}x + c \cr & \therefore \,{\sin ^{ - 1}}y - {\sin ^{ - 1}}x = c \cr} $$

$$\eqalign{ & {\text{Given }}\frac{{dy}}{{dx}} = 1 + {y^2} \cr & \Rightarrow \frac{{dy}}{{1 + {y^2}}} = dx \cr & \Rightarrow {\tan ^{ - 1}}y = x + c \cr & \Rightarrow y = \tan \left( {x + c} \right) \cr & {\text{Now }}y\left( 0 \right) = 0 \Rightarrow \tan \,c = 0 \cr & y\left( \pi \right) = 0 \Rightarrow \tan \left( {\pi + c} \right) = 0 \Rightarrow c = n\pi \cr & \therefore \,\,y = \tan \,x \cr} $$

$$\eqalign{ & \phi \left( x \right)dy = y\phi \left( x \right)dx - {y^2}dx \cr & {\text{or }}\frac{{y\phi '\left( x \right)dx - \phi \left( x \right)dy}}{{{y^2}}}dy - dx = 0 \cr & {\text{or }}d\left( {\frac{{\phi \left( x \right)}}{y}} \right) - dx = 0 \cr & {\text{or }}d\left( {\frac{{\phi \left( x \right)}}{y} - x} \right) = 0 \cr} $$

Given $$\frac{{x\,dx}}{{1 + {x^2}}} = \frac{{y\,dy}}{{1 + {y^2}}}$$
Integrating we get,
$$\eqalign{ & \frac{1}{2}\log \left( {1 + {x^2}} \right) = \frac{1}{2}\log \left( {1 + {y^2}} \right) + a \cr & \Rightarrow 1 + {x^2} = c\left( {1 + {y^2}} \right),\,c = {e^{2a}} \cr & {x^2} - c{y^2} = c - 1 \Rightarrow \frac{{{x^2}}}{{c - 1}} - \frac{{{y^2}}}{{\left( {\frac{{c - 1}}{c}} \right)}} = 1......\left( 1 \right) \cr} $$
Clearly $$c > 0{\text{ as }}c = {e^{2a}}$$
Hence, the equation $$\left( 1 \right)$$ gives a family of hyperbolas with eccentricity
$$ = \sqrt {\frac{{c - 1 + \frac{{c - 1}}{c}}}{{c - 1}}} = \sqrt {\frac{{{c^2} - 1}}{{c - 1}}} = \sqrt {c + 1} {\text{ if }}c \ne 1$$
Thus eccentricity varies from member to member of the family as it depends on $$c.$$ If $$c = 1,$$  it is a pair of lines $${x^2} - {y^2} = 0$$

Let the slope of tangent of required family be $$\frac{{dy}}{{dx}} = {m_1}$$
Also $$y = \frac{{{C^2}}}{x}\,;$$   therefore, $$\frac{{dy}}{{dx}} = - \frac{{{C^2}}}{{{x^2}}} = {m^2}\,\,\,\left( {{\text{say}}} \right)$$
By the given condition, we have
$$\eqalign{ & \tan \frac{\pi }{4} = \frac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}} \cr & \Rightarrow 1 + {m_1}{m_2} = {m_1} - {m_2} \cr & \Rightarrow \frac{{dy}}{{dx}} + \frac{{{C^2}}}{{{x^2}}} = 1 - \frac{{{C^2}}}{{{x^2}}}\frac{{dy}}{{dx}} \cr & \Rightarrow \frac{{dy}}{{dx}}\left( {1 + \frac{{{C^2}}}{{{x^2}}}} \right) = 1 - \frac{{{C^2}}}{{{x^2}}} \cr & \Rightarrow \frac{{dy}}{{dx}} = \frac{{{x^2} - {C^2}}}{{{x^2} + {C^2}}} \cr} $$

$$\frac{{dx}}{{dy}} + \frac{x}{y} - {y^2} = 0\,;\,\frac{{dx}}{{dy}} + \frac{x}{y} = {y^2}$$
This is a linear differential equation of the form
$$\eqalign{ & \frac{{dx}}{{dy}} + {P_1}x = {Q_1}\,; \cr & {\text{Here, }}P = \frac{1}{y}{\text{ and }}Q = {y^2} \cr & \therefore {\text{ I}}{\text{.F}}{\text{.}} = {e^{\int {P\,dy} }} = {e^{\int {\frac{1}{y}\,dy} }} = {e^{\log \,y}} = y \cr} $$
So, required solution is
$$\eqalign{ & x.y = \int {{y^2}.y\,dy + c\,;\,xy} = \int {{y^3}dy + c} \cr & xy = \frac{{{y^4}}}{4} + c\,;\,4xy = {y^4} + c \cr} $$