Assume $z=f(x,y)$ with $x= \rho \cos(\theta)$ and $y= \rho \sin(\theta)$,then show
$$(\frac{\partial z}{\partial x})^2+(\frac{\partial z}{\partial y})^2=(\frac{\partial z}{\partial \rho})^2+\frac{1}{\rho^2}(\frac{\partial z}{\partial \theta})^2$$
$$\frac{\partial z}{\partial \theta}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial \theta}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial \theta}$$
and $$\frac{\partial z}{\partial \rho}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial \rho}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial \rho}$$
But I don't know how to compute $\frac{\partial z}{\partial x}$ or $\frac{\partial z}{\partial y}$,can someone help me?
Just write down the expressions:
$$\left(z_\rho\right)^2=\left(z_xx_\rho+z_yy_\rho\right)^2=\left(z_x\cos t+z_y\sin t\right)^2=z_x^2\cos^2t+2z_xz_y\cos t\sin t+z_y^2\sin^2t$$
and
$$\frac1{\rho^2}(z_t)^2=\frac1{\rho^2}\left(z_x(-\rho\sin t)+z_y(\rho\cos t)\right)^2=$$
$$=\frac1{\rho^2}\left(z_x^2\rho^2\sin^2 t-2z_xz_y\rho^2\cos t\sin t+z_y^2\rho^2\cos^2t\right)$$
Well, now just add the above two expressions and see what you get...