Help with differentiability of $\frac{x^6}{x^4 + y^4}$ at $(x,y)=(1,2)$

Question

Imagine you have the following function: $$ f(x, y): \mathbb R^2 \rightarrow \mathbb R \\ \; \\ f(x) = \begin{cases} \frac{x^6}{x^4 + y^4} & (x, y) \neq (0, 0) \; \; \\ \\ 0 & (x, y) \neq (0, 0) \\ \end{cases} $$ And I'm trying to prove that it is differentiable in $(x,y) = (1, 2)$ $$ \lim_{t \to 0} \frac{\frac{(1 + t)^6}{(1 + t)^4 + (2 + t)^4} - f(\mathbf x_0) - \frac{\partial f}{\partial \mathbf x}(1, 2) \cdot(t,t) }{t} $$ Computing we get the following: $$ \lim_{t \to 0} \frac{(1 + t)^6 - f(\mathbf x_0) [(1 + t)^4 + (2 + t)^4] - t\frac{\partial f}{\partial x}(1, 2) - t\frac{\partial f}{\partial y}(1, 2) }{t[(1 + t)^4 + (2 + t)^4]} $$ We know that the partial derivatives exists, and the function value, also exists at $(1, 2)$. But the thing is that this limit is undefined, but indeed the function is differentiable at that point, so, what I'm doing wrong? How would I compute this limit?

Answer 1

Compute the partial derivatives: $$\frac{\partial }{\partial x}\left(\frac{x^6}{x^4 + y^4}\right)=\frac{2x^9+6x^5y^4}{(x^4 + y^4)^2} $$ $$\frac{\partial }{\partial y}\left(\frac{x^6}{x^4 + y^4}\right)=-\frac{4y^3x^6}{(x^4 + y^4)^2} $$ Now compute the gradient at the point $(1,2)$: $$\nabla f(1,2)=\left(\frac{98}{289},-\frac{128}{289} \right)$$ Finally compute the following limit to see if $f$ is differentiable: $$\lim_{(x,y)\to (1,2)}\frac{1}{\sqrt{x^2+y^2}}\left(f(x,y)-f(1,2)-\langle\nabla{f}(1,2),(x-1,y-2)\rangle \right) $$ $$\lim_{(x,y)\to (1,2)}\frac{1}{\sqrt{x^2+y^2}}\left(\frac{x^6}{x^4 + y^4}-\frac{1}{17}-\frac{98}{289}(x-1) +\frac{128}{289} (y-2)\right)=0 $$ Hence the function is differentiable at $(x,y)=(1,2)$.