Meaningless partial derivative 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)
 A: 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.
A: I think your confusion is when you said about the limit
$$\lim_{t\to0}\frac{f(t,0)-f(0,0)}{t}=\lim_{t\to 0}\frac{0}{t}$$
that it is not really meaningful. Note that in this limit "$t\to0$" means that $t$ approaches zero but is not equal to zero. So for each admissible value of $t$, the expression $\displaystyle\frac{0}{t}$ is perfectly well-defined, and is in fact equal to zero — which makes the limit perfectly meaningful and in fact equal to $0$ as well.
