Solution: The only problem is to understand whether or not this function is differentiable at point $(0,0)$. If the given function is differentiable at $(0,0)$, then it must be linear and hence all the directional derivatives will also be linear. Recall that the directional derivative at a point $\mathbf{x} $ in the direction of $\mathbf{v} $ is given by \[ df_{\mathbf{x} }(\mathbf{v} ) = \lim_{t \to 0} \frac{f(\mathbf{x} + t\mathbf{v} )-f(\mathbf{x} )}{t}. \]

If $\mathbf{v} = (v_1,v_2)$, then the directional derivative of the given function at $(0,0)$ will be \begin{align*} df_{(0,0)}(\mathbf{v} ) & = \lim_{t \to 0} \frac{f(t \mathbf{v} ) - f(0,0)}{t} \\[1ex] & = \lim_{t \to 0} \frac{\frac{t^3 v_1^3 + v_2^4 t^4}{t^2 v_1^2 + t^2 v_2^2}-0}{t} \\[2ex] & = \lim_{t \to 0} \frac{v_1^3 + tv_2^4}{v_1^2 + v_2^2} \\[1ex] & = \frac{v_1^3}{v_1^2 + v_2^2}, \end{align*} which is not linear. Hence, the given function is not differentiable at $0$.

Another way to tackle this problem is to see the linear approximation of $f(x,y)$ at $(0,0)$. The linear approximation of the function is given by \[ L(x,y) = f(0,0) + x\cdot\frac{\partial f}{\partial x}(0,0)+ y\cdot \frac{\partial f}{\partial y} (0,0). \] Note that using the directional derivative computation, we have \begin{align*} \frac{\partial f}{\partial x}(0,0) = df_{(0,0)}(1,0) = 1 \\ \frac{\partial f}{\partial y}(0,0) = df_{(0,0)}(0,1) = 0. \end{align*} Therefore, the linear approximation will be \[ L(x,y) = x. \]

For $f$ to be differentiable at $(0,0)$ we would need the limit \[ \lim_{(x,y) \to (0,0)} \frac{f(x,y)-L(x,y)}{\sqrt{x^2 +y^2} } \] to be equal to $0$. Note that \begin{align*} \lim_{(x,y) \to (0,0)} \frac{f(x,y)-L(x,y)}{\sqrt{x^2 +y^2} } & = \lim_{(x,y) \to (0,0)} \frac{\frac{x^3 + y^4}{x^2 + y^2}-x}{\sqrt{x^2 +y^2} } \\[1ex] & = \lim_{(x,y) \to (0,0)} \frac{x^3 + y^4 -x^3 -y^2x}{(x^2 + y^2)^{\frac{3}{2}}} \\[1ex] & = \lim_{(x,y) \to (0,0)} \frac{y^4 - y^2x}{(x^2 +y^2)^{\frac{3}{2}}}. \end{align*}

Note that the above limit does not exist. For that take the limit along the line $y=mx$. Then we have \begin{align*} \lim_{(x,y) \to (0,0)} \frac{y^4 - y^2x}{(x^2 +y^2)^{\frac{3}{2}}} & = \lim_{x \to 0} \frac{m^4x^4 - m^2 x^3}{\left( x^2 + m^2 x^2 \right) ^{\frac{3}{2}}} \\[1ex] & = \lim_{x \to 0} \frac{x^3m^2 \left( xm^2 - 1 \right) }{x^3 \left( 1+m^2 \right) ^{\frac{3}{2}}} \\[1ex] & = \frac{-m^2}{ \left( m^2 + 1 \right) ^{\frac{3}{2}}}. \end{align*} The above limit depends upon $m$ and hence the limit does not exist as the limit along two different lines are different. Therefore, the function is not differentiable at $(0,0)$.