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Let the function $f$ be defined as $$f(x,y)=\frac{x^4-y^3}{x^2+y^2}$$ If $(x,y)=(0,0)$ then $f$ is equal to zero.
My problem is to prove that this function isn't differentiable at the point $(0,0)$.

My solution:
First idea is maybe to show that $f$ is not continuous at $(0,0)$. That will show that $f$ isn't differentiable at $(0,0)$.
But it is easy to see that $f$ is continuous at $(0,0)$, so we need to check the partial derivatives at $(0,0)$.
Partial derivative in $x$-direction is $$\lim_{h\to 0}\frac{f(h,0)-0}{h}=0$$
Partial derivative in $y$-direction is $$\lim_{h\to 0}\frac{f(0,h)-0}{h}=-1$$ We now know how our Jacobian matrix at the point $(0,0)$ looks. We now use the definition of derivatives: $$\lim_{(x,y)\rightarrow(0,0)}\frac{\frac{x^4-y^3}{x^2+y^2}+y}{\sqrt{(x^2+y^2)}}=\frac{x^4+x^2y}{(x^2+y^2)^{\frac{3}{2}}}$$ But this looks like having a limit in $(0,0)$ and it's equal to $0$?

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We consider the function \begin{align*} f(x,y)=\begin{cases} \frac{x^4-y^3}{x^2+y^2}&\qquad\text{if }(x,y)\ne (0,0)\\ 0&\qquad\text{if }(x,y)=(0,0) \end{cases} \end{align*}

As already stated by OP we see the partial derivatives exist. \begin{align*} \frac{\partial f}{\partial x}(0,0)&=\lim_{h\rightarrow 0}\frac{f(h,0)-f(0,0)}{h} =\lim_{h\rightarrow 0}\frac{h^4}{h^3}=0\\ \frac{\partial f}{\partial y}(0,0)&=\lim_{h\rightarrow 0}\frac{f(0,h)-f(0,0)}{h} =\lim_{h\rightarrow 0}\frac{-h^3}{h^3}=-1\\ \end{align*}

We observe that if the function $f$ is differentiable at $(0,0)$ the linear approximation $l(x,y)$ must be given as \begin{align*} l(x,y)&=f(0,0)+\frac{\partial f}{\partial x}(0,0)\cdot x+\frac{\partial f}{\partial y}(0,0)\cdot y\\ &=-y\tag{1} \end{align*}

For $f$ to be differentiable at $(0,0)$ we would need the limit \begin{align*} \lim_{(x,y)\rightarrow(0,0)}\frac{f(x,y)-l(x,y)}{\sqrt{x^2+y^2}} \end{align*} to be equal to $0$ due to (1). Here the limit is \begin{align*} \lim_{(x,y)\rightarrow(0,0)}\frac{x^4-y^3}{(x^2+y^2)^{\frac{3}{2}}} \end{align*}

If this limit exists, it must be equal to the limit as $(x,y)$ approaches the origin along the line $y=x$, which would be \begin{align*} \lim_{x\rightarrow 0}\frac{x^4-x^3}{(2x^2)^{\frac{3}{2}}} =\frac{1}{2\sqrt{2}}\lim_{x\rightarrow 0}\frac{x^4-x^3}{(x^2)^{\frac{3}{2}}}\tag{2} \end{align*}

Since \begin{align*} \lim_{x\rightarrow 0^+}\frac{x^4-x^3}{(x^2)^{\frac{3}{2}}}=-1\qquad\qquad\text{resp.}\qquad\qquad \lim_{x\rightarrow 0^-}\frac{x^4-x^3}{(x^2)^{\frac{3}{2}}}=1\\ \end{align*} the limit (2) is not equal to $0$ showing the function $f$ is not differentiable at $(0,0)$.

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    $\begingroup$ @LovroSindicic:Thanks a lot for accepting my answer and granting the bounty! :-) $\endgroup$
    – epi163sqrt
    Feb 1 '17 at 11:32
  • $\begingroup$ Why is $\lim_{(x,y)\rightarrow(0,0)}\frac{f(x,y)-l(x,y)}{\sqrt{x^2+y^2}} = \lim_{(x,y)\rightarrow(0,0)}\frac{x^4-y^3}{(x^2+y^2)^{\frac{3}{2}}}$? Shouldn't it be equal to $\lim_{(x,y)\rightarrow(0,0)}\frac{x^4-y^3 +y}{(x^2+y^2)^{\frac{3}{2}}}$ arccording to $(1)$? $\endgroup$ Jun 17 '17 at 22:15
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    $\begingroup$ @ReluxtheRelux: Thanks for your hint. I will check and if necessary revise my answer soon. $\endgroup$
    – epi163sqrt
    Jun 18 '17 at 16:39
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A necessary condition for being differentiable at $(0,0)$ is that the directional derivative: $$ Df_{(0,0)}(h,k) = \lim_{t\rightarrow 0} \frac{1}{t} (f(th,tk)-f(0,0)) = \lim_{t\rightarrow 0} \frac{t^4 h^4 - t^3 k^3}{t^3(h^2+k^2)} = \frac{-k^3}{h^2+k^2}$$ should be a linear function of $h$ and $k$ (i.e. of the form $ah+bk$ for real constants $a$ and $b$), which is visibly not the case. In your own solution you may simply calculate $Df_{(0,0)} (1,1)$ and show that it is not the sum of $Df_{(0,0)} (1,0)$ and $Df_{(0,0)} (0,1)$, so it is not differentiable.

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Let $Q(x,y):=\frac{x^4+x^2y}{(x^2+y^2)^{\frac{3}{2}}}$. Then show that for $y=x>0$ we have

$Q(x,x)=\frac{1}{\sqrt{8}}(x+1) \to \frac{1}{\sqrt{8}} \ne 0$ for $x \to 0$.

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  • $\begingroup$ Why this prove that the llimit is $\frac{1}{\sqrt8}$? $\endgroup$
    – josf
    Jan 24 '17 at 10:27
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    $\begingroup$ By definition: $f$ is differentiable at $(0,0)$ iff $Q(x,y) \to 0$ for $(x,y) \to (0,0)$. From $Q(x,x)=\frac{1}{\sqrt{8}}(x+1) \to \frac{1}{\sqrt{8}} \ne 0$ for $x \to 0+$, we see that $Q(x,y)$ does not converge to $0$ for $(x,y) \to (0,0)$. Thus: $f$ is not differentiable at $(0,0)$ $\endgroup$
    – Fred
    Jan 24 '17 at 12:22
  • $\begingroup$ Ok, I see now. I see higher power of x in numerator than in denominator so I think that limit could equist. $\endgroup$
    – josf
    Jan 24 '17 at 13:12

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