Show that the partial derivatives of $x^3/{x^2+y^2}$ are bounded. Find if they are continuous.
I calculated the partial derivatives to be $${\partial f\over\partial x}={x^2(3y^2+x^2)\over(x^2+y^2)^2}$$ $${\partial f\over\partial y}={-2x^3y\over(x^2+y^2)^2}$$
But I'm not sure how to show that they are bounded and/or continuous
Assuming that your calculations are correct, you can simply divide top and bottom by $x^4$ to obtain
$$\frac{\partial f}{\partial x} = \frac{x^2(3y^2-x^2)}{(x^2+y^2)^2} = \frac{3r^2-1}{(1+r^2)^2}$$
$$\frac{\partial f}{\partial y} = \frac{-2x^3y}{(x^2+y^2)^2} = \frac{-2r}{(1+r^2)^2}$$
where $r = \frac yx$. Then prove that the RHS is bounded for $r \in [0,\infty)$ by differentiating with respect to $r$ and finding the extremum.