Difference between Gâteaux and Frêchet derivatives I can't seem to grasp the difference between the definitions of gâteaux and Frêchet derivatives I am wokring with:
A function $f: U \rightarrow \mathbb{R}^{n}$ ($U$ open) is said to be Gâteaux diff. in $x_0$ if every directional derivative $D_vf(x_0)$ exists ($\forall v$) and the map $v \mapsto D_vf(x_0)$ is linear.
A function is said to be Frêchet diff. in $x_0$ if a linear map $L:\mathbb{R}^n \rightarrow \mathbb{R}$ such that $$\lim_{x \to x_0} \frac{f(x)-f(x_0)-L(x-x_0)}{|x-x_0|} = 0$$
Now by my understanding, as we work over $\mathbb{R}^{n}$, $L$ is equal to the total derivative $df(x_0)$. If a function is frêchet diff., we have $df(x_0) = \langle\nabla f(x_0), \cdot\rangle$, I understand that if a function is frêchet diff in $x_{0}$, it is also gâteaux diff in $x_0$ as exist (whereas $v\mapsto \langle \nabla f(x_0), v\rangle$)
But how does the map $v \mapsto D_vf(x_0)$ from the gâteaux definition differ from the total derivative $df(x_0)$ ? I mean the map in the gâteaux definition is also linear so what is difference between both?
I know there are similar questions out there, but I could not find one that used the same definitions.
 A: Best to have some examples where it is possible to calculate things. In this one, we switch to polar coordinates to see that it is Gateaux differentiable, indeed all directional derivatives are zero. However, it is not Frechet differentiable, as a path that is not a straight line gives a poorly behaved quotient.  
Take $$ g(0,0) = 0, \; \mbox{otherwise} \; \; g(x,y) = \frac{x^5 \; y^5}{x^{12} + y^8}   $$
Added, August 20, 2017: it occurred to me that I had not proved $g$ continuous. Lagrange multipliers says that the maximum of $|x^5 y^5|,$ for fixed $x^{12} + y^8,$ occurs when $2 y^8 = 3 x^{12}.$ Fiddle with it a little, we get
$$ |g| \leq C \; \; \sqrt[\color{red}{24}] {x^{12} + y^8},  $$ where
$$ C = \frac{ 2^{\left( \frac{5}{12}\right)}  3^{\left( \frac{5}{8}\right)}}{  5^{\left( \frac{25}{24}\right)}}.$$
So $g$ is continuous at the origin.  
Next, we write the function itself in polar coordinates,
$$ g = r^2 \left( \frac{\cos^5 \theta \; \; \sin^5  \theta}{r^4 \cos^{12} \theta + \sin^8 \theta} \right) $$
Along either coordinate axis, we get constant zero. Otherwise, with the function defined as zero at the origin, for Gateaux we simply take $g/r,$ giving
$$ \frac{g}{r} \; = \; \; r \; \; \left( \frac{\cos^5 \theta \; \; \sin^5  \theta}{r^4 \cos^{12} \theta + \sin^8 \theta} \right) $$
For any fixed $\theta$ direction (not on the x,y axes), this gives limit $0$ as $r \rightarrow 0,$ therefore  directional derivative zero.
Everything looks promising in polar coordinates, giving Gateaux.
But then, along path $x = t^2, y = t^3,$ we find the quotient
$$ \frac{g(t^2, t^3)}{\sqrt{t^4 + t^6}} \geq \frac{1}{4|t|} $$
when $|t| \leq \sqrt 3,$ so $g$ is not Frechet.
