We have

$\dfrac {\partial x}{\partial u} = \cos v, \ \dfrac {\partial x}{\partial v} = - u \sin v$

And $\dfrac {\partial y}{\partial u} = \sin v, \ \dfrac {\partial y}{\partial v} = u \cos v$

hence,

$\dfrac {\partial (x, y)}{\partial (u, v)} = J = \begin{vmatrix} x_u &x_v \\y_u &y_v \end{vmatrix} = \begin{vmatrix} \cos v &-u \sin v \\\sin v &u \cos v \end{vmatrix}$

$\therefore J = u \cos^2 v + u \sin ^2 v = u$

$\therefore \dfrac {\partial (x, y)}{\partial (u,v)} = u$

Also, $\tan v= y/x $ i.e. $v =\tan ^{-1} y/x$

$\therefore \dfrac {\partial v}{\partial x} = \dfrac {-y}{x^2 + y^2} \text { and } \dfrac {\partial v}{\partial y} = \dfrac {x}{x^2 + y^2} \text{ and } \cos^2 v + \sin^2 v =1 $

$\therefore x^2 + y^2 = u^2$

$\therefore u \dfrac {\partial u}{\partial x} = x \text{ and } v \dfrac {\partial u}{\partial y} = y$

$J' = \begin{vmatrix} u_x &u_v \\v_x &v_y \end{vmatrix} = \begin{vmatrix} \frac {x}{u} & \frac {y}{u} \\ \frac {-y} {x^2 + y^2} & \frac {x}{x^2 + y^2} \end{vmatrix} = \dfrac {1}{u}$

$\therefore JJ' = 1 \text{ hence } \dfrac {\partial (x,y)}{\partial( u,v)} \cdot \dfrac {\partial (u,v)}{\partial (x,y)} = 1$