$ux+vy=a$ Differentiating partially w.r.t u, treating v as constant, $x+u \left ( \dfrac {\partial x}{\partial u} \right )_v + v \left ( \dfrac {\partial y}{\partial u}\right )_v=0 $ $\dfrac {u}{x}+ \dfrac {v}{y}= 1$ Differentiating partially w.r.t u, treating v as constant, $\dfrac {x-u \left ( \frac {\partial x}{\partial u} \right )_v}{x^2}- \dfrac {v}{y^2}\left ( \dfrac {\partial y}{\partial u} \right )_v=0$ $Eliminating \ \left ( \dfrac {\partial y}{\partial u} \right )_v $ $x+u \left ( \dfrac {\partial x}{\partial u} \right )_v+ \dfrac {y^2}{x^2 } \left ( x-u \left ( \dfrac {\partial x}{\partial u} \right )_v \right ) = 0 $ $\left ( \dfrac {\partial x}{\partial u} \right )_v \left [ux^2 - y^2 u \right ]= -x^3 - y^2 x $ $\left (\dfrac {\partial x}{\partial u} \right )_v = \dfrac{x (x^2 +y^2)}{u (x^2-y^2)}$ $\dfrac {u}{x}\left (\dfrac {\partial x}{\partial u} \right )_v = - \dfrac {x^2 + y^2}{x^2 -y^2}$ Similarly, we get $\dfrac {v}{y} \left ( \dfrac {\partial y}{\partial v} \right )_u = \dfrac {x^2+y^2}{x^2 -y^2} $ Adding them, $\dfrac {u}{x} \left ( \dfrac {\partial x}{\partial u} \right )_v + \dfrac {v}{y}\left ( \dfrac {\partial y}{\partial v} \right )_u =0$