Suppose a function $f$ is defined as follows:
$$f(x,y)=\begin{cases} \frac{x^3y^2}{x^4+y^4}&\text{ when }(x,y)\neq(0,0),\\0 & \text{ when }(x,y)=(0,0).\end{cases}$$
I want to determine whether or not the function is Fréchet differentiable at $(0,0)$? I don't think that it is.
Here's my reasoning: if $f$ is Fréchet differentiable at $(0,0)$, then from the definition of Fréchet differentiability, the Fréchet derivative must be f itself. But this is not a linear map ($f(1,0)=0$ and $f(0,1)=0$ but $f(1,1) \neq 0$). By contradiction, f is not Fréchet differentiable at $(0,0)$.
Is this correct?
You got all you need at an answer to your earlier question... put in polar,
$$\frac{x^3y^2}{x^4+y^4}= r\,\, \left(\frac{\cos^3(\theta)\sin^2(\theta)}{\cos^4(\theta)+\sin^4(\theta)}\right)$$
This says that your function is continuous, because the quantity in the parentheses has some maximum absolute value which you could find with calculus.
It also says that the function is Gateaux differentiable at the origin, and tells you the directional derivative for any (unit length) direction.
Finally, it says the function cannot be Frechet differentiable, because the directional derivatives along the x,y axes are zero; these are the partial derivatives. If Frechet differentiable, linerarity would demand that the directional derivative in every direction be zero, but that is not the case, just plug in $\theta = \pi / 4$ to get something nonzero. Notice that David C. Ulrich suggested you look at $(t,t),$ same thing.
Extra credit: even if the directional derivatives of some complicated function are linear, for example all zero at some point, that still does not guarantee Frechet differentiability. Explicit bounds are needed for the linear approximation. There are many answers about this distinction, here is mine: Directional derivatives in any direction are all equal with example
Take $$ f(0,0) = 0, \; \mbox{otherwise} \; \; f(x,y) = \frac{x^{12/5} \; y^{6/5}}{x^4 + y^2} $$ If you are uncomfortable with the fractional exponents for numbers that may be negative, switch to $$ f(0,0) = 0, \; \mbox{otherwise} \; \; f(x,y) = \frac{|x|^{12/5} \; |y|^{6/5}}{x^4 + y^2} $$ which is the same.
EDIT:
found one with integer exponents,
Take $$ g(0,0) = 0, \; \mbox{otherwise} \; \; g(x,y) = \frac{x^5 \; y^5}{x^{12} + y^8} $$ Everything looks promising in polar coordinates, but then $$ \frac{|g(t^2, t^3)|}{\sqrt{t^4 + t^6}} \geq \frac{1}{4|t|} $$ when $t \neq 0$ and $|t| \leq \sqrt 3.$