My problem is that I get two different answers when I use two different approaches to differentiate this problem with respect to $x$:
$$\left(\frac{3x+2y^2}{x^2+y^2}\right) =4$$
My first thought was to use the quotient rule, which gives me the answer
$$\frac{dy}{dx} = \frac{4x^2y+3x^2-3y^2}{4x^2y-6xy}.$$
However, someone showed me that if you first get rid of the rational expression like this:
$$3x+2y^2 = 4x^2+4y^2$$
and then differentiate, you get
$$\frac{dy}{dx}=\frac{8x-3}{4y}$$ (which is also the answer wolfram alpha gives).
I tried this also on a much simpler rational expression (to check that the problem wasn't just me beigng bad at algebra) and got different results again! Why?
Thanks for any help!
Good news: the results may seem different, but aren't.
There's a small (sign) mistake here, you should find: $$\frac{dy}{dx}=\frac{3-8x}{4y}$$
This may look different from the form you found via the quotient rule, but remember that you have a relation between $x$ and $y$, the original (implicit) function. So we hope the following equality holds:
$$\begin{array}{crcl} {} & \displaystyle \frac{4x^2y+3x^2-3y^2}{4x^2y-6xy} & = & \displaystyle \frac{3-8x}{4y} \\[5pt] {\Leftrightarrow} & 4y\left( 4x^2y+3x^2-3y^2 \right) & = & \left( 4x^2y-6xy \right) \left( 3-8x \right)\end{array}$$
Expanding, moving everything to the left-hand side and simplifying gives:
$$2 y \left( 16 x^3-24 x^2+4 x \cdot \color{blue}{2y^2}+9 x-3 \cdot \color{blue}{2y^2} \right)=0 \tag{$*$}$$
But from: $$\left(\frac{3x+2y^2}{x^2+y^2}\right) =4$$ we have that: $$3x+2y^2=4x^2+4y^2 \Rightarrow \color{blue}{2y^2=3x-4x^2}$$
Substitution into $(*)$ and simplifying, will give you $0$.