If $${x^a}{y^b} = {\left( {x - y} \right)^{a + b}},$$     then the value of $$\frac{{dy}}{{dx}} - \frac{y}{x}$$   is equal to :

Solution :
$$\eqalign{ & {x^a}{y^b} = {\left( {x - y} \right)^{a + b}}{\text{ taking }}\log {\text{ both sides}}{\text{.}} \cr & \log \left( {{x^a}{y^b}} \right) = \log {\left( {x - y} \right)^{\left( {a + b} \right)}} \cr & a\log \,x + b\,\log \,y = \left( {a + b} \right)\log \left( {x - y} \right) \cr & {\text{differentiating both sides w}}{\text{.r}}{\text{.t}}{\text{. }}'x' \cr & \frac{a}{x} + \frac{b}{y}\frac{{dy}}{{dx}} = \frac{{\left( {a + b} \right)}}{{\left( {x - y} \right)}}\left[ {1 - \frac{{dy}}{{dx}}} \right] \cr & \frac{{dy}}{{dx}}\left[ {\frac{b}{y} + \frac{{a + b}}{{x - y}}} \right] = \frac{{a + b}}{{x - y}} - \frac{a}{x} \cr & \frac{{dy}}{{dx}}\left[ {\frac{{bx - by + ay + by}}{{y\left( {x - y} \right)}}} \right] = \frac{{ax + bx - ax + ay}}{{x\left( {x - y} \right)}} \cr & \frac{{dy}}{{dx}}\left[ {\frac{{bx + ay}}{y}} \right] = \frac{{bx + ay}}{x} \cr & \frac{{dy}}{{dx}} = \frac{y}{x}\,;\,\,\frac{{dy}}{{dx}} - \frac{y}{x} = 0 \cr} $$