$v(x,y), u(x,y)$ are differentiable. $v_x=-u_y, u_x=v_y$. Given $x=r\cos\theta, y=r\sin\theta$ $v(x,y), u(x,y)$ are differentiable. $v_x=-u_y, u_x=v_y$. Given $x=r\cos\theta, y=r\sin\theta$, Prove that for every $(x,y)\neq(0,0) : v_r=\frac{-1}{r}u_\theta, u_r=\frac{1}{r}v_\theta$.

I'm clueless in this case. How should I approach this type of question?
by the chain rule :

$ \begin{align} u_r = u_x x_r+u_yy_r = u_x \cos \theta +u_y \sin \theta \\
   v_r = v_x x_r+v_yy_r = v_x \cos \theta +v_y \sin \theta \\
u_{\theta} = u_x x_{\theta}+u_yy_{\theta} = -rv_y\sin \theta -rv_x \cos\theta = -rv_r \implies v_r=\frac{-1}{r}u_{\theta} \\ 
v_{\theta} = v_x x_{\theta}+v_yy_{\theta} = ru_y\sin \theta +ru_x \cos\theta = ru_r \implies u_r=\frac{1}{r}v_{\theta}
\end{align} $
we have, (dv/dx)=−(du/dy) and (du/dx)=(dv/dy)
'derivative of v with respect to x= product of derivative of r with respect to x and derivative of v with respect to r' as well as in y and θ

given, x=rcosθ,y=rsinθ => r=x/cosθ=y/sinθ and cosθ=(x/r)and sinθ=(y/r)
dr/dx=1/cosθ, dr/dy=1/sinθ and sinθ.dθ=-(1/r).dx, cosθ.dθ= (1/r).dy

(dv/dx)=−(du/dy) =>(dr/dx)(dv/dr)=−(dθ/dy)(du/dθ) =>(1/cosθ)(dv/dr)=-(1/(r.cosθ)).(du/dθ)=>(dv/dr)=-(1/r).(du/dθ)

(du/dx)=(dv/dy)=>(dr/dx)(du/dr)=(dθ/dy)(dv/dθ) =>(1/sinθ)(dv/dr)=(1/(r.sinθ)).(du/dθ)=>(du/dr)=(1/r).(dv/dθ)