<s>[INST] Partial derivative of $f(x,y) = \frac{x^3+y^3}{x^2+y^2}$ at $f_1(0,0)$ To be more precise than the title, the function is actually piecewise
$$
f(x,y) = \begin{cases}
\frac{x^3+y^3}{x^2+y^2} & (x,y) \ne (0,0) \\
0 & (x,y) = (0,0) \\
\end{cases}
$$
I checked that the function is continuous at $(0,0)$, so I then calculated the partial derivative with respect to $x$ as
$$
f_1(x,y) = \frac{x^4-2xy^3+3x^2y^2}{(x^2+y^2)^2} \tag{1}
$$
This is undefined at $(0,0)$, so I then tried to find the limit around accumulation points. Let $S_1$ be the points on the $x$ axis
$$
\lim_{x \to 0} f_1(x,0) = \frac{x^4}{(x^2)^2} = 1 \tag{2}
$$
Let $S_2$ be the points on the line $y = x$
$$
\lim_{x \to 0} f_1(x,x) = \frac{x^4-2x^4+3x^4}{(x^2+x^2)^2} = \frac{2x^4}{(2x^2)^2} = \frac{1}{2} \tag{3}
$$
So, the limits are different around different accumulation points. That's where I'm confused because the answer should be $1$.
 [/INST] Hint: Compute the limit (as $h$ goes to $0$) of  $(f(0+h,0)-f(0,0))/h$ to get the value of the partial derivative at $(0,0)$. 
What you are looking at with those path limits is the question of continuity of this partial derivative function at $(0,0)$. 
</s>