Existence of the limit of $\lim_{(x,y) \to (0,0)}(y\ln(xy)+y)$ Let be $x$ and $y$ positive, $f(x,y)=xy\ln(xy)$.
The limit of $\lim_{(x,y) \to (0,0)}f(x,y)$ exists, I used following steps:
\begin{equation}
\lim_{(x,y) \to (0,0)}(xy\ln(xy)) = \lim_{(x,y) \to (0,0)}(\ln(xy)^{xy})= \ln(1)=0
\end{equation}
To calculate $\lim_{(x,y) \to (0,0)} f'_x(x,y)$ I did following steps:
\begin{equation}
f'_x(x,y)= y\ln(xy)+y
\end{equation}
\begin{equation}
\lim_{(x,y) \to (0,0)}(y\ln(xy)+y) = \lim_{(x,y) \to (0,0)}(\ln(xy)^y+y)= \ln(1) + 0=0
\end{equation}
The solution is that the second limit doesn't exist. What did I wrong?
Is the first solution with $f(x,y)$ right?
 A: I'm going to assume you mean $f(x,y) = xy\ln(xy)$.
In your first displayed equation, you seem to be missing some parentheses, and you seem to assert  $\lim_{(x,y)\to(0,0)} \ln((xy)^{xy})=\ln(1)=0$.  This is correct because it can be proven that $t^t \to 1$ as $t \to 0^+$.
The limit of $f_x(x,y) \equiv \frac{\partial f}{\partial x}(x,y) = y\ln(xy) + y$ as $(x,y) \to (0,0)$ doesn't exist  ($f'_x$ is bad notation).  Hints:
$y \to 0$ as $(x,y) \to (0,0)$.
$\ln(xy) = \ln x + \ln y$ for $x, y > 0$.
It is well-known that $y\ln y \to 0$ as $y\to 0^+$.  You probably know how to do this already.
That leaves $y\ln x$, where $y$ and $x$ are both close to $0$.  $\ln x \to -\infty$ as $x \to 0^+$.  You should be able to find a path in the $x>0, y>0$ quadrant of the $xy$-plane which goes to $(0,0)$, along which $y\ln x$ approaches any non-positive value you please, including $-\infty$, or maybe doesn't even exist at all.
Your mistake in the "$f_x$" limit was in claiming that $\lim_{(x,y)\to (0,0)}(xy)^y$ is $1$ (again, you left out some important parentheses).  The problem is that unlike in the first problem, the base and the exponent are different, $xy$ and $y$ can go to $0$ at different rates, so $(xy)^y$ can approach $0$, $1$, anything in between, or might not converge to any limit at all as $(x,y)$ approaches the origin.
A: Note that $y\ln(xy)=y\ln{x}+y\ln{y}.$ As you show, 
$\lim\limits_{(x,y) \to (0,0)}{y\ln{y}}=0,$ but
$\lim\limits_{(x,y) \to (0,0)}{y\ln{x}}$ does not exist. To prove this, we consider two curves $\gamma_1$ and $\gamma_2:$
$$\gamma_1=\{x(t),\  y_1(t),\;\; 0<t<+\infty \},\\
  \gamma_2=\{x(t),\  y_2(t),\;\; 0<t<+\infty \},$$
where
$$x(t)=\dfrac{1}{e^t},\;\;  y_1(t)= \dfrac{1}{t},\;\; y_2(t)= \dfrac{2}{t}.$$
We have $x(t)\to{0}, \;\; y_1(t) \to {0}, \;\;y_2(t) \to {0}$ as $t\to +\infty.$ But
$$\lim\limits_\underset{(x,y)\in{\gamma_{1}}}{(x,y) \to (0,0)}{y\ln{x}}=\lim\limits_{t\to +\infty}{\dfrac{-t}{t}}=-1,\\
\lim\limits_\underset{(x,y)\in{\gamma_{2}}}{(x,y) \to (0,0)}{y\ln{x}}=\lim\limits_{t\to +\infty}{\dfrac{-2t}{t}}=-2.
$$
A: The term $h(x,y)=y\ln(xy)$ has no limit when $(x,y)\to(0,0)$ since $h(x,x)=2x\ln x\to0$ while $h(\mathrm e^{-1/y},y)\to-1$ when $y\to0^+$.
A: I'll consider your function $f(x,y)=xy\ln(xy)$ defined only for $x>0$, $y>0$. One can write
$$
f(x,y)=xy\ln x+xy\ln y.
$$
Now, when $0<x<1$, we have $|xy\ln y|\le |y\ln y|$ and, as $\lim_{y\to0}y\ln y=0$, we have
$$
\lim_{(x,y)\to(0,0)}xy\ln y=0.
$$
Similarly for the limit of the other summand. Therefore
$$
\lim_{(x,y)\to(0,0)}f(x,y)=0.
$$
Consider now $f'_x(x,y)=y\ln(xy)+y=y\ln y+y+y\ln x$. The limit of this function at $(0,0)$ doesn't exist: its existence would imply that of
$$
\lim_{(x,y)\to(0,0)}x^y
$$
which is known not to exist.
A: That is not correct. The limit does not exist. In particular,
$$
\lim_{(x,y)\to (0,0)}y\ln y+y=0,
$$
but 
$$
\lim_{(x,y)\to (0,0)}y\ln x \quad \text{does not exist!}
$$
To see this, take $x=y$, then $x\ln x\to 0$, as $x\to 0$. But if we take $x=\mathrm{e}^{-1/y^2}$, then
$$
y\ln x=y\ln \mathrm{e}^{-1/y^2}=-\frac{1}{y}\to-\infty,
$$
as $y\to 0^+$.
