differentiability of $(x^2-y)y^2/(x^4+y^2)$?

Question

$f(x,y)=(x^2-y)y^2/(x^4+y^2)$ for every $(x,y)\neq (0,0)$ and $f(x,y)=(0,0)$ on $(0,0)$. I am trying to find out whether this function is differentiable or not. I proved that its continuous on the origin and I have calculated the definition of derivative (using limit) for different paths but I could not prove that the function is not differentiable so I am guessing that it is but I do not know how to prove it.

Answer 1

I am going to introduce the definition of differentiability of a function. A function $f : \mathbb{R}^n \to \mathbb{R}^m$ is differenctiable in $x_0\in \mathbb{R}^n$ if there exists a linear function $L:\mathbb{R}^n\to\mathbb{R}^m$ such that: $$ \lim_{h\to (0,0)}\frac{\left|\left|\,f(x_0+h)-f(x_0)-L(h)\right|\right|}{\left|\left|h\right|\right|}=0 $$ The matrix associated to the linear function $L$ is the Jacobian matrix.

Then, let's calculate the Jacobian matrix of: $$ f(x,y)=\begin{cases} \displaystyle\frac{(x^2-y)y^2}{x^4+y^2} & \text{if } (x,y)\neq (0,0) \\ 0 & \text{if } (x,y)=(0,0) \end{cases} $$ in $(x,y)=(0,0)$. First: $$ \frac{\partial f}{\partial x}(0,0) = \lim_{h\to 0} \frac{0}{h^4}=0 $$ $$ \frac{\partial f}{\partial y}(0,0) = \lim_{h\to 0} \frac{-h^3}{h^2}=0 $$

Then the Jacobian matrix in $(0,0)$ of $f$ is $J_{(0,0)}(\,f)=\left(\begin{array}{cc} 0 & 0 \end{array}\right)$, the if $L$ exists, it is $L(x,y)=\left(\begin{array}{cc} 0 & 0 \end{array}\right)\left(\begin{array}{c} x \\ y \\ \end{array}\right)=0$. Then:

\begin{eqnarray} \lim_{(x,y)\to (0,0)}\frac{\left|\left|\,f(x,y)-f(0,0)-L(x,y)\right|\right|}{\left|\left|(x,y)\right|\right|} &=& \lim_{(x,y)\to (0,0)}\frac{\left|\left|\,f(x,y)\right|\right|}{\left|\left|(x,y)\right|\right|} \\ \lim_{(x,y)\to (0,0)}\frac{\left|\left|\,f(x,y)-f(0,0)-L(x,y)\right|\right|}{\left|\left|(x,y)\right|\right|} &=& \lim_{(x,y)\to (0,0)}\left|\frac{(x^2-y)y^2}{x^4+y^2}\,\right|\frac{1}{\left|\left|(x,y)\right|\right|} \\ \end{eqnarray}

If we use the path $y=x$: \begin{eqnarray} \lim_{(x,y)\to (0,0)}\frac{\left|\left|\,f(x,y)-f(0,0)-L(x,y)\right|\right|}{\left|\left|(x,y)\right|\right|} &=& \lim_{x\to 0}\left|\frac{(x^2-x)x^2}{x^4+x^2}\,\right|\frac{1}{\left|\left|(x,x)\right|\right|} \\ \lim_{(x,y)\to (0,0)}\frac{\left|\left|\,f(x,y)-f(0,0)-L(x,y)\right|\right|}{\left|\left|(x,y)\right|\right|} &=& \lim_{x\to 0}\left|\frac{x^2-x}{x^2+1}\,\right|\frac{1}{|x|\left|\left|(1,1)\right|\right|} \\ \lim_{(x,y)\to (0,0)}\frac{\left|\left|\,f(x,y)-f(0,0)-L(x,y)\right|\right|}{\left|\left|(x,y)\right|\right|} &=& \frac{1}{\left|\left|(1,1)\right|\right|} \lim_{x\to 0}\left|\frac{x-1}{x^2+1}\,\right| \\ \lim_{(x,y)\to (0,0)}\frac{\left|\left|\,f(x,y)-f(0,0)-L(x,y)\right|\right|}{\left|\left|(x,y)\right|\right|} &=& \frac{1}{\left|\left|(1,1)\right|\right|} \neq 0 \\ \end{eqnarray} Thus $f$ is not differentiable in $(0,0)$