Directional derivative of a two dimensional piecewise function

Question

Let $f:\mathbb{R^2} \rightarrow \mathbb{R}$ with $f(x,y)= \begin{cases} \frac{x^4y}{x^4+y^4} & (x,y)\neq(0,0) \\ 0 & (x,y)=(0,0) \end{cases}$

Show that $f$ is differentiable in all directions $v=(a,b)$. Is $f$ differentiable in $(0,0)$?

I'm a little confused by the question possibly because I'm wrong about the definition of directional derivative but we define a directional derivative as: $$D_vf(x_0,y_0)=v \cdot \nabla f(x_0,y_0)= v \cdot \begin{pmatrix} \partial_xf(x_0,y_0) \\ \partial_yf(x_0,y_0) \end {pmatrix} $$ Applying the definition we have $$D_vf(x,y)=v\cdot \begin {pmatrix} \frac{4x^3y^5}{(x^4+y^4)^2} \\\frac{x^4(x^4-3y^4)}{(x^4+y^4)^2} \end {pmatrix} $$ However unless I made a big mistake the limits of the individual partial derivatives don't converge to zero so the function shouldn't be differentiable, so how does the directional derivative exist? Is there a problem with what I've done so far or is the statement wrong?

Answer 1

HINT

To tackle the first part, it is advisable to apply the definition: \begin{align*} D_{v}f((0,0)) & = \lim_{t\to 0}\frac{f((0,0) + tv) - f(0,0))}{t} = \lim_{t\to 0}\frac{f(ta,tb)}{t} = \lim_{t\to 0}\frac{t^{5}a^{4}b}{t^{5}a^{4} + t^{5}b^{4}} = \frac{a^{4}b}{a^{4} + b^{4}} \end{align*} and the desired result holds.

Can you handle the second part?