I am confused about the relationship between Frechet differentiable and strong Fréchet differentiable.
Assume the function $f(x) \in \mathbb{R}, x\in \mathbb{R}^n$ is Fréchet differentiable at $x$.
Then we have $$\lim_{x \neq y^1 \to x} \frac{f(y^1) - f(x) - \nabla f(x)^T(y^1-x)}{\Vert y^1-x\Vert} = 0$$
$$\lim_{x \neq y^2 \to x} \frac{f(y^2) - f(x) - \nabla f(x)^T(y^2-x)}{\Vert y^2-x\Vert} = 0$$
Then use the first equation minus the second one and the inequality $\Vert y^1-x\Vert \leq \Vert y^1-y^2\Vert + \Vert y^2 - x\Vert$, then we get
$$\lim_{\substack{y^1 \neq y^2 \\ (y^1, y^2) \to (x, x)}} \frac{f(y^1) - f(y^2) - \nabla f(x)^T(y^1-y^2)}{\Vert y^1 - y^2 \Vert} = 0, $$ which is the definition of strong F-derivative.
It seems I did not use any other condition but F-derivative to get strong F-derivative. I am quite confused about their difference.
The typical counterexample is $f \colon \mathbb R \to \mathbb R$, $$f(x) = \begin{cases} 0 & \text{if } x = 0, \\ x^2 \sin(x^{-1}) & \text{if } x \ne 0. \end{cases}$$ This function is Fréchet differentiable on all of $\mathbb R$, but not strictly/strongly Fréchet differentiable at $x = 0$.
Since $f'(0) = 0$, it is also very instructive (and not too hard) to use this counterexample and see where your argument fails.
If I remember correctly, one even has the following: If the function $f$ is Fréchet differentiable everywhere, then the strict/strong differentiability is equivalent to the continuity of the derivative.