Answer
$$\lim _{(x, y) \rightarrow(0,0)} \frac{x+y}{x^2+y} $$
does not exist
Work Step by Step
Given$$\lim _{(x, y) \rightarrow(0,0)} \frac{x+y}{x^2+y} $$
So, $$\lim _{(x, y) \rightarrow(0,0)} \frac{0+0}{0^2+0} =\frac{0}{0}$$
so, we get along the line $x=0$
\begin{align}
L&=\lim _{(x, y) \rightarrow(0,0)} \frac{x+y}{x^2+y} \\
&=\lim _{(0, y) \rightarrow(0,0)} \frac{y }{y } \\
&=1
\end{align}
And therefore we get along the line $y=0$
\begin{align}
L&=\lim _{(x, y) \rightarrow(0,0)} \frac{x+y}{x^2+y} \\
&=\lim _{(x, 0) \rightarrow(0,0)} \frac{x }{x^2 } \\
&=\lim _{(x, 0) \rightarrow(0,0)} \frac{1}{x } \\
&=\infty\\
\end{align}
So, the limit does not exist
