Elementary Second partial derivative If $V=\frac{xy} {(x^2+y^2)^2}$ and $x=r\cos\theta$, $y=r\sin\theta$ show that $\frac{\partial^2V} {\partial{r^2}}+\frac{1}{r}\frac{\partial{V}} {\partial{r}}+\frac{1} {r^2} \frac{\partial^2{V}}{\partial{\theta^2}}=0$
My attempt :
$\frac{\partial {V}} {\partial{x}} =y\cdot\frac{y^2-3x^2} {(x^2+y^2)^3}\\$
$\frac{\partial {V}} {\partial{y}} =x\cdot\frac{y^2-3x^2} {(x^2+y^2)^3}$
$\frac{\partial{V} } {\partial{r}}= \frac{\partial {V}} {\partial{x}}\frac{\partial {x}} {\partial{r}}+\frac{\partial {V}} {\partial{y}}\cdot\frac{\partial {y}} {\partial{r}}\\\frac{\partial {V}} {\partial{r}}=\frac{y^2-3x^2} {(x^2+y^2)^3}\{y\cos\theta+x\sin\theta\}$
$\frac{\partial^2{V} } {\partial{r}^2} =\frac{\partial} {\partial{r}}\biggr\{\frac{y^2-3x^2} {(x^2+y^2)^3}\{y\cos\theta+x\sin\theta\} \biggr\}\\=$
$\frac{\partial} {\partial{\theta}}\biggr\{\frac{y^2-3x^2} {(x^2+y^2)^3}\{y\cos\theta+x\sin\theta\} \biggr\} \frac{\partial{\theta} }{\partial{r}}$
At this point it is getting clumsy. Any hint on this.

Hint: Write $$V=\frac{\sin2\theta}{2r^2}.$$Now take partial derivative with respect to $r$ and $\theta$.
