Trigonometry Derivative ('Show That' Question)

Q: Given that $\displaystyle y= \mathrm{arctan} \frac{x}{a}$, where $\displaystyle a$ is a constant, show that $\displaystyle x \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\tan y}{1 + \tan ^2 y}$.

So... I have done (and reached a dead end):

$\displaystyle y= \mathrm{arctan} \frac{x}{a}$

$\displaystyle \therefore \tan y = \frac {x}{a}$

$\displaystyle \frac{\tan y}{1 + \tan ^2 y}$

$\displaystyle = \frac{(\frac{x}{a})}{1 + (\frac{x}{a})^2}$

$\displaystyle = \frac{(\frac{x}{a})}{\frac{a^2 + x^2}{a^2}}$

$\displaystyle =\frac{ax}{a^2 + x^2}$

But That doesn't lead me anywhere. Please help. Thanks in advance.