Find $dy/dx$ of $(xy^2)+5 = x + 2y^2$ For the solution I got $$\frac{y^2-1}{ 4y-2xy} = dy/dx$$
I just want to know if this is correct. Also it says to evaluate $dy/dx$ at $(1,2)$.
Would the solution to that be $3/4$?

Here are the steps
$$ xy^2+5=x+2y^2 $$
$$ \frac{d}{dx}\left[xy^2+5\right]=\frac{d}{dx}\left[x+2y^2\right] $$
$$ y^2\frac{d}{dx}\left[x\right]+x\frac{d}{dx}\left[y^2\right]=1+2\frac{d}{dx}\left[y^2\right] $$
$$ y^2+2xy\frac{d}{dx}[y]=1+4y\frac{d}{dx}[y] $$
$$ y^2-1=\left(4y-2xy\right)\frac{d}{dx}[y] $$
$$ \frac{d}{dx}[y] = \frac{y^2-1}{4y-2xy} $$
So now at $(1, 2)$, we have
$$ \frac{d}{dx}[y] = \frac{2^2-1}{4\cdot 2-2\cdot 1\cdot 2} = \frac{3}{8-4} = \frac34  $$
