If $h(r, \theta) = f(r \cos \theta, r \sin \theta)$, show that $ f_{xx}+f_{yy} = h_{rr} + \frac{1}{r} h_r + \frac{1}{r^2} h_{00}$ 
If $h(r, \theta) = f(r \cos \theta, r \sin \theta)$, show that 
  $$ f_{xx}+f_{yy} = h_{rr} + \frac{1}{r} h_r + \frac{1}{r^2} h_{00}$$
  Hint: Rewrite the defining equation as $f(x,y) = h(r(x,y), \theta(x,y))$, with $r(x,y) = \sqrt{x^2+y^2}$ and $\theta (x,y) = tan^{-1} (\frac{y}{x})$, and differentiate with respect to $x$ and $y$.

Defining equation as $$f(x,y) = h \left[r(x,y), \theta(x,y)  \right]$$, with $$r(x,y) = \sqrt{x^2+y^2} \\ \theta (x,y) = tan^{-1} (\frac{y}{x})$$
(1) 
$$f_x = \frac{\partial h }{\partial r } \frac{\partial r }{\partial x  } +\frac{\partial  h }{\partial \theta } \frac{\partial \theta  }{\partial x } = 
\frac{\partial h }{\partial r } \frac{y }{(y^2+x^2)^{\frac{1}{2} } } - \frac{\partial  h }{\partial \theta } \frac{y}{x^2+y^2} \tag{A}
$$
$$f_{xx} = \frac{\partial}{\partial x} \left( \frac{\partial h }{\partial r } \frac{y }{\sqrt{y^2+x^2} } - \frac{\partial  h }{\partial \theta } \frac{y}{x^2+y^2}  \right)  = \frac{\partial^2 h }{\partial x \partial r } \frac{y^2 }{(y^2+x^2)^{\frac{3}{2}} } - \frac{\partial^2  h }{\partial x \partial \theta } \frac{2yx}{(x^2+y^2)^2} \tag{B}
$$
.
Also from tag A, substituting $\left(  \frac{\partial h}{\partial r}      \right)$ for $(h)$
$$\frac{\partial}{\partial x}(h) = \frac{\partial}{\partial r } (h) \frac{y }{(y^2+x^2)^{\frac{1}{2} } } - \frac{\partial   }{\partial \theta } (h) \frac{y}{x^2+y^2}$$
$$ \implies \frac{\partial^2 h }{\partial x \partial r } = \frac{\partial}{\partial x} \left(  \frac{\partial h}{\partial r}      \right) = \frac{\partial}{\partial r } \left(  \frac{\partial h}{\partial r}      \right) \frac{y }{(y^2+x^2)^{\frac{1}{2} } } - \frac{\partial   }{\partial \theta } \left(  \frac{\partial h}{\partial r}      \right) \frac{y}{x^2+y^2} $$ $$
\implies  \frac{\partial^2 h }{\partial x \partial r }=
 \left(  \frac{\partial^2 h}{\partial r^2}      \right) \frac{y }{(y^2+x^2)^{\frac{1}{2} } } -  \left(  \frac{\partial^2 h}{\partial \theta .\partial r}      \right) \frac{y}{x^2+y^2}   \tag{C} $$
Second Substitution
$$ \implies \frac{\partial^2  h }{\partial x \partial \theta } =  \left(  \frac{\partial^2 h}{\partial r \partial \theta}      \right) \frac{y }{(y^2+x^2)^{\frac{1}{2} } } -  \left(  \frac{\partial^2 h}{\partial \theta^2.}      \right) \frac{y}{x^2+y^2} \tag{D} $$
It follows, placing (C) and (D) in (B):
$$f_{xx} =  \left[      \left(  \frac{\partial^2 h}{\partial r^2}      \right) \frac{y }{(y^2+x^2)^{\frac{1}{2} } } -  \left(  \frac{\partial^2 h}{\partial \theta .\partial r}      \right) \frac{y}{x^2+y^2}       \right]  \frac{y^2 }{(y^2+x^2)^{\frac{3}{2}} } - \left(  \left(  \frac{\partial^2 h}{\partial r \partial \theta}      \right) \frac{y }{(y^2+x^2)^{\frac{1}{2} } } -  \left(  \frac{\partial^2 h}{\partial \theta^2.}      \right) \frac{y}{x^2+y^2}  \right) \frac{2yx}{(x^2+y^2)^2}
$$
.
(2) I would then proceed similarly with $f_y$
$$f_y = \frac{\partial h }{\partial r } \frac{\partial r }{\partial y } +\frac{\partial  h }{\partial \theta } \frac{\partial \theta  }{\partial y } = 
\frac{\partial h }{\partial r } \frac{y }{(y^2+x^2) ^{\frac{1}{2}}       }+\frac{\partial  h }{\partial \theta } \frac{x}{x^2+y^2} 
$$
...
(3) After Adding $f_{xx}$ and $f_{yy}$ , and simplifying, Hopefully, I should get an expression such as $h_{rr} + \frac{1}{r} h_r + \frac{1}{r^2} h_{00}$.
.
.
The rest of my result does not seem to lead to the desired outcome $h_{rr} + \frac{1}{r} h_r + \frac{1}{r^2} h_{00}$; there is no $h_r$ to be seen in $f_{xx}+f_{yy}$.
EDIT- Is the approach, eventhough long, I undertook would normally let one able to show  that $h_{rr} + \frac{1}{r} h_r + \frac{1}{r^2} h_{00}$? Is there an alternative approach to the one of Mr. Surd shown as answer involving only manipulation of partials (The chapter where the problem is does not have any concept related to the one shown by Mr. Surd).
Much appreciated for your input/help.

Let $f(x,y)=f(r\cos\theta ,r\sin\theta )=h(r,\theta ).$
Then, $$\nabla _{(r,\theta )}h=\begin{pmatrix}\cos\theta &\sin\theta \\ -r\sin\theta &r\cos\theta \end{pmatrix}\nabla f\implies \nabla f=\frac{1}{r}\begin{pmatrix}r\cos\theta &-\sin\theta \\r\sin\theta &\cos\theta \end{pmatrix}\nabla _{(r,\theta )}h=:\begin{pmatrix}h_1(r,\theta )\\ h_2(r,\theta )\end{pmatrix}.$$
Now, you can observe that $$\Delta f=\nabla \cdot \nabla f,$$ and thus, you get $$\Delta f=\cos\theta \frac{\partial h_1}{\partial r}-\frac{\sin\theta}{r} \frac{\partial h_1}{\partial \theta }+\sin\theta \frac{\partial h_2}{\partial r}+\frac{\cos\theta}{r} \frac{\partial h_2}{\partial \theta }.$$
I let you finishing all the calculation.
