So here is the question:
$$ \tan^{-1}\left(\frac{2x}{y}\right)=\frac{\pi x}{y^2} $$
When I solved it implicitly I got (with much pain in formatting it on this site :P):
$$ y^2=\pi \left(\frac{y^2-2xy\cdot y'}{y^4}\right)\cdot \left ( \sec^2\left(\frac{2x}{y}\right)\cdot (2y-2xy') \right ) $$
Now I know this sounds stupid but I don't know how to factor out y' because apparently I have derived correctly to the best of my knowledge and yet when I input (1,2) in my function and then check Wolphram Alpha, I get two different results (that shouldn't be the case)....
I'm at a loss as to what to do... Any help would be appreciated
Write your problem as $F[x,y]=0$ and perform total differentiation.
So $$0 = dF = (dF/dx) dx + (dF/dy) dy$$ and then $$y' = \frac{dy}{dx} = - \frac{dF/dx}{dF/dy}$$