The function in question is
$$f(x,y) = \begin{cases} 0 & (x,y)=(0,0)\\ \frac{xy}{|x|+|y|} & (x,y) \neq (0,0) \end{cases} $$
So far I've calculated the continuity of the function at $(0,0)$ using the squeeze theorem with the functions $xy$ and $\frac{xy}{\sqrt{x^2+y^2}}$. The former is always bigger than the second branch of $f$, and the latter always smaller; they all have limit $0$, therefore the function is continuous at $(0,0)$.
Now I have to calculate the first partial derivative in order to $x$ of $f$ at $(0,0)$. Using the definition, we have
$$\frac{\partial f}{\partial x}(0,0)=\lim_{t\to 0}\frac{f(t,0)-f(0,0)}{t}$$
Now, $f(0,0)$ is just $0$, and $f(t,0)$ is $0$ as well. So we have
$$\lim_{t\to 0}\frac{f(t,0)-f(0,0)}{t}=\lim_{t\to 0}\frac{0}{t}$$
which is not really meaningful, at least to me, and there doesn't seem to be an obvious way of solving this limit without finding the full expression.
So I started hunting for an expression for the partial derivative using the limit definition. That should be
$$\frac{\partial f}{\partial x}(x_0,y_0)=\lim_{t\to 0}\frac{f(x_0+t,y_0)-f(x_0,y_0)}{t}$$
but that seems impressingly difficult to calculate. If any of this is correct, we have
$$\lim_{t\to 0}\frac{\frac{(x+t)y}{|x+t|+|y|}-\frac{xy}{|x|+|y|}}{t}=\lim_{t\to 0}\frac{(x+t)y}{t(|x+t|+|y|)}-\frac{xy}{t(|x|+|y|)}$$
and after a few "simplifications" (which don't simplify anything at all), we get
$$\lim_{t\to 0} \frac{xy|x|+ty|x|+xy|y|+ty|y|-xy|x+t|-xy|y|}{t(|x+t|+|y|)(|x|+|y|)}$$
and there's where it becomes a huge mess. I think separating $|x+t|$ would help a lot, but I have no idea on how to justify doing that. This analysis class is killing me, even if you can't help me with the problem just recommending a good textbook would help a ton.
Here's the function plotted on WA: https://www.wolframalpha.com/input/?i=f(x,y)+%3D+(xy)%2F(%7Cx%7C%2B%7Cy%7C)
Very roughly speaking, the partial derivative $\frac{\partial f}{\partial x}$ represents the change in the value of $f$ in response to a very small change in the value of $x$, assuming that $y$ is fixed. Note that if we fix $y = 0$, then $$ f(x,0) = \frac{x\cdot 0}{|x|+|0|} = 0. $$ Hence if $y=0$, the function is constant with respect to $x$ (and equal to zero). Thus a small change in $x$ will lead to no change in the value of $f$. Therefore it is reasonable to expect that the partial derivative with respect to $x$ will be zero.
More formally, $$ \frac{\partial f}{\partial x}(x,0) = \lim_{t\to 0} \frac{f(t,0) - f(0,0)}{t} = \lim_{t\to 0} \frac{0}{t} = 0, $$ which is what we (heuristically) expect.