Is a function $C^1$ on $\mathbb{R}^2$?

Partial derivatives away from the origin. For $(x,y) \neq (0,0)$,
$$\frac{\partial f}{\partial x} = \frac{(3x^2 y)(x^2+y^2) - (x^3 y)(2x)}{(x^2+y^2)^2} = \frac{x^2 y(x^2+3y^2)}{(x^2+y^2)^2}.$$
Similarly,
$$\frac{\partial f}{\partial y} = \frac{x^3(x^2+y^2) - x^3 y(2y)}{(x^2+y^2)^2} = \frac{x^3(x^2-y^2)}{(x^2+y^2)^2}.$$

Partial derivatives at $(0,0)$. By definition,
$$\frac{\partial f}{\partial x}(0,0) = \lim_{h\to0} \frac{f(h,0)-f(0,0)}{h} = \lim_{h\to0} \frac{0}{h} = 0.$$
Similarly, $\frac{\partial f}{\partial y}(0,0) = 0$.

Continuity at $(0,0)$. Using $|x| \le \|(x,y)\|$ and $|y| \le \|(x,y)\|$,
$$\left|\frac{\partial f}{\partial x}(x,y)\right| \le \frac{\|(x,y)\|^3 \cdot \|(x,y)\| \cdot 4\|(x,y)\|^2}{\|(x,y)\|^4} = 4\|(x,y)\| \to 0.$$
So $\frac{\partial f}{\partial x}$ is continuous at $(0,0)$, and similarly for $\frac{\partial f}{\partial y}$.

Conclusion. The partial derivatives exist and are continuous on $\mathbb{R}^2$, so $f$ is $C^1$.