The limit is $0$, but we have to prove it using the epsilon-delta definition. How do we make the epsilon-delta relation for this?
I found that the limits for the function itself is $0$ by finding the limits on $y$ axis( where $x=0$), on $x$ axis( where $y=0$), when $y=x$, when $y=x^3$, where the limit was always $0$ therefore the limit is $0$.
As noted in A. P.'s comment, the limit being zero along any given path is necessary but not sufficient. If you want to prove this in general, use inequalities, in particular that $x^2 + 1 \geq 1$ for all $x$ and thus $\left|\frac{x+y}{x^2+1}\right| \leq \left| x+y \right|$ for all $x,y$.