A little guidance on finding the limit How do I find the limit of $f(z) = \frac{x^2y}{x^3+y^3} + ixy$  as $z \to0$ ?
What I think is if $z\to0$, that implies $x ,y\to0$. But since the $f(z)$ has both variables $x$ and $y$ mixed together, it will always lead to $0$ no matter which values approach $0$ first.
To me, It seems just not right. Are there any general rules to follow when finding the limit of a function?
PS: My mathematics background is poor so please use simple explanations. 
 A: Here are some additional informations similar to that of @freak_warrior.

Important aspect: Limits are unique
  
  
*
  
*We have to keep in mind, that if a limit exists it is unique.
  
  
  This implies that independent of the path we use to approach to zero, we always have to get the same value.
  
  
*
  
*On the other hand, if we can show  that along two different pathes we get two different values,  we can immediately conclude that the limit does not exist and the job is done.
  

We want to analyse
\begin{align*}
\lim_{z\rightarrow 0}f(z)=\lim_{x+iy\rightarrow 0}\left(\frac{x^2y}{x^3+y^3}+ixy\right)
\end{align*}
We try do make the task simple by first checking simple pathes and look, if they already lead to different values when we approach to zero along them. Simple pathes are e.g. straight lines. And simple straight lines (in terms of simple mathematical description) are the $x$-axis and the $y$-axis. 

  
*
  
*First path: $x$-axis
  
  
  The $x$-axis is described by $y=0$. The function $f$ along the $x$-axis is given as
$$f(x+i0)=\frac{x^2\cdot0}{x^3+0^3}+ix\cdot 0=0$$
So, walking along the $x$-axis gives always $0$ and we get
$$\lim_{x\rightarrow 0}f(x+i0)=0$$
Conclusion: If a limit $\lim_{z\rightarrow 0}f(z)$ exists, it has to be $0$, since a limit is unique.

Let's check another simple path.

  
*
  
*Second path: $y$-axis
  
  
  The $y$-axis is described by: $x=0$. The function $f$ along the $y$-axis is given as
$$f(0+iy)=\frac{0^2\cdot y}{0^3+y^3}+i 0\cdot y=0$$
So, walking along the $y$-axis gives also always $0$ and we get
$$\lim_{y\rightarrow 0}f(0+i\cdot y)=0$$

Now a third trial. We test the diagonal $x=y$

  
*
  
*Third path: the diagonal $x=y$
  
  
  The function $f$ along the diagonal $x=y$ is given as
$$f(x+ix)=\frac{x^2\cdot x}{x^3+x^3}+ix^2=\frac{x^3}{2x^3}+ix^2=\frac{1}{2}+ix^2$$
So, walking along the diagonal $x=y$ and approaching to $0$ gives
$$\lim_{x\rightarrow 0}f(x+i\cdot x)=\lim_{x\rightarrow 0}\left(\frac{1}{2}+ix^2\right)=\frac{1}{2}$$
Voilà, we found a different value when approaching to zero via the main diagonal.



Conclusion: Since the limit along the $x$-axis is different from the limit along the main diagonal $x=y$, the limit
  $$\lim_{z\rightarrow0}f(z)$$
  does not exist.


Comments:


*

*One counterexample is sufficient in order to show that the limit $\lim_{z\rightarrow 0}f(z)$ does not exist.

*Even if the limit $f(z)$ exists along along all different pathes, this is not sufficient for the existence of the limit $\lim_{z\rightarrow 0}f(z)$. They all have additionally to be of the same value.

*We could find a counterexample, by finding polynomials of the same degree in numerator ($x^2\cdot x$) and denominator ($x^3+x^3$). This is often helpful in case we have examples with fractions involved.
A: "What I think is if $z\to0$, that implies $x ,y\to0$. ". This part is correct.
However, for your case here, $x ,y$ can tend to $0$ along different paths. 
For example, along $y=x$ and $y=-x$ [HINT]. So you need check whether these 2 paths would lead to different values. 
