Solution: We recall that \begin{align*} \frac{\partial f}{\partial x} (a,b) = \lim_{h \to 0} \frac{f(a + h, b) - f(a,b)}{h} \\[2ex] \frac{\partial f}{\partial y} (a,b) = \lim_{k \to 0} \frac{f(a, b + k) - f(a,b)}{k}. \end{align*}

Note the function is continuous and differentiable everywhere other that $(0,0)$. We need to check at $(0,0)$. So all the partial derivatives exist at $(a,b)$ if $(a,b) \neq (0,0)$. Let us compute the partial derivatives at $(0,0)$. \begin{align*} \frac{\partial f}{\partial y} (0,0) & = \lim_{k \to 0} \frac{f(0, k) - f(0,0)}{k} \\ & = \lim_{k \to 0} \frac{0 - 0}{k^3} = 0. \end{align*} Similarly, \begin{align*} \frac{\partial f}{\partial x} (0,0) & = \lim_{h \to 0} \frac{f(0, k) - f(0,0)}{h} \\ & = \lim_{h \to 0} \frac{0 - 0}{h^2} = 0. \end{align*}

For other points, we can compute the partial derivative as usual \begin{align*} \frac{\partial f}{\partial x} & = \begin{cases} \dfrac{\partial }{\partial x} \left( \frac{xy^2}{x + y^2} \right) , &\text{ if } (x,y) \neq (0,0);\\ 0, &\text{ if } (x,y) = (0,0). \end{cases} \\[2ex] & = \begin{cases} \dfrac{y^2(x + y^2) - xy^2 (1)}{((x + y^2)^2}, &\text{ if } (x,y) \neq (0,0);\\ 0, &\text{ if } (x,y) = (0,0). \end{cases} \\[2ex] & = \begin{cases} \dfrac{y^4}{(x + y^2)^2}, &\text{ if } (x,y) \neq (0,0) ;\\ 0, &\text{ if } (x,y) = (0,0). \end{cases} \end{align*} Similarly, \begin{align*} \frac{\partial f}{\partial y} & = \begin{cases} \dfrac{\partial }{\partial y} \left( \frac{xy^2}{x + y^2} \right) , &\text{ if } (x,y) \neq (0,0);\\ 0, &\text{ if } (x,y) = (0,0). \end{cases} \\[2ex] & = \begin{cases} \dfrac{2xy(x + y^2) - xy^2 (2y)}{((x + y^2)^2}, &\text{ if } (x,y) \neq (0,0);\\ 0, &\text{ if } (x,y) = (0,0). \end{cases} \\[2ex] & = \begin{cases} \dfrac{2x^2 y}{(x + y^2)^2}, &\text{ if } (x,y) \neq (0,0);\\ 0, &\text{ if } (x,y) = (0,0). \end{cases} \end{align*} Now we have two functions namely $\frac{\partial f}{\partial x} $ and $\frac{\partial f}{\partial y} $.

We need to compute the second derivative at $(0,0)$. We can use the same methods that we used before. \begin{align*} \frac{\partial^2 f}{\partial x \partial y} (0,0) & = \left.\frac{\partial }{\partial x}\right|_{(0,0)} \left( \frac{\partial f}{\partial y} \right) \\ & = \lim_{h \to 0} \frac{\frac{\partial f}{\partial y} (h,0) - \frac{\partial f}{\partial y} (0,0)}{h} \\[1ex] & = \lim_{h \to 0} \frac{0}{h^3} = 0 \end{align*} Similarly, \begin{align*} \frac{\partial^2 f}{\partial y \partial x} (0,0) & = \left.\frac{\partial }{\partial y}\right|_{(0,0)} \left( \frac{\partial f}{\partial x} \right) \\ & = \lim_{k \to 0} \frac{\frac{\partial f}{\partial x} (0,k) - \frac{\partial f}{\partial x} (0,0)}{k} \\[1ex] & = \lim_{k \to 0} \frac{0}{k^5} = 0 \end{align*} Thus, \[ \textcolor{teal}{\boxed{ \frac{\partial^2 f}{\partial x \partial y}(0,0) = 0 \text{ and } \frac{\partial^2 f}{\partial y \partial x}(0,0) = 0. }} \]

Now to discuss the continuity of the second derivative, we first need to compute at a nonzero point similar to the earlier case. \begin{align*} \frac{\partial^2 f}{\partial x \partial y} & = \begin{cases} \dfrac{\partial }{\partial x} \left( \frac{2x^2 y}{(x + y^2)^2} \right) , &\text{ if } (x,y) = (0,0);\\ 0, &\text{ if } (x,y) = (0,0). \end{cases} \\[1ex] & = \begin{cases} \dfrac{4xy^3}{(x + y^2)^3}, &\text{ if } (x,y) \neq (0,0);\\ 0, &\text{ if } (0,0).) \end{cases} \end{align*}

To talk about the continuity of the function $\frac{\partial^2 f}{\partial x \partial y} $, we take a path \[ x = r^2 \cos ^2 \theta , \text{ and } y = r \sin \theta , \ \ r \geq 0. \] If the function is continuous, the limit as $r$ approaches to zero, must be zero. That is, \begin{align*} \lim_{r \to 0} \frac{\partial^2 f}{\partial x \partial y}(x,y ) & = \lim_{r \to 0} \frac{4 r^5 \cos ^2\theta \sin ^3\theta }{(r^2\cos ^2\theta +r^2\sin ^2\theta )^3} \\ & = \lim_{r \to 0} \frac{4 r^5 \cos ^2 \theta \sin ^3\theta }{r^6} \rightarrow \infty . \end{align*} Thus, it is not continuous at $(0,0)$.

Similarly, \begin{align*} \frac{\partial^2 f}{\partial y \partial x} & = \begin{cases} \dfrac{\partial }{\partial y} \left( \frac{y^4}{(x + y^2)^2} \right) , &\text{ if } (x,y) = (0,0);\\ 0, &\text{ if } (x,y) = (0,0). \end{cases} \\[1ex] & = \begin{cases} \dfrac{4xy^3}{(x + y^2)^3}, &\text{ if } (x,y) \neq (0,0);\\ 0, &\text{ if } (x,y) = (0,0). \end{cases} \end{align*} Since $\frac{\partial^2 f}{\partial y \partial x} = \frac{\partial^2 f}{\partial x \partial y}$, and $\frac{\partial^2 f}{\partial x \partial y}$ is not continuous at $(0,0)$, so is the function $\frac{\partial^2 f}{\partial y \partial x}$.