How do I use the definition of partial derivative to get $f_x(x,y) = \frac{y(y^2-x^2)}{(x^2 + y^2)^2}$? $\\f(x,y) =
\begin{cases}
\frac{xy}{x^2 + y^2},  & \text{if $(x,y) \ne (0,0)$} \\
0, & \text{if $(x,y) = (0,0)$}
\end{cases}$
Using the definition of partial derivatives: 
\begin{align*}
    f_x(x,y) &= \lim_{h\to 0} \frac{f(x+h,y) - f(x,y)}{h}
    \\&= \lim_{h\to 0} \frac{\frac{(x+h)y}{(x+h)^2 + y^2}-\frac{xy}{x^2 + y^2}}{h}\\&= \lim_{h\to 0} \frac{(xy + hy)(x^2 +y^2)-xy(x^2 +2hx+ h^2 + y^2)}{h(x^2 + 2hx + h^2 + y^2)(x^2 + y^2)}
    \end{align*}
I'm not sure how to get rid of the variable $h$. I wanted to know exactly how I could further simplify in order to get $\frac{y(y^2-x^2)}{(x^2 + y^2)^2}$?

Continuing your calculation, we get
$$f_x(x,y)=\lim_{h\rightarrow 0}\frac{hy(y^2-x^2-hx)}{h(x^2+2hx+h^2+y^2)(x^2+y^2)}.$$
Or
$$f_x(x,y)=\lim_{h\rightarrow 0}\frac{y(y^2-x^2-hx)}{(x^2+2hx+h^2+y^2)(x^2+y^2)}.$$
Can you finish from here?
