Frechet Differentiability versus Strict Differentiability We recall two concepts of differentiability of a mapping. Let $X, Y$ be real Banach spaces and $f: X\rightarrow Y$
be a mapping. 


$f$ is said to be Frechet differentiable at $\bar{x}$ if there
exists a linear continuous operator $\nabla f(\bar{x}): X\rightarrow Y$ such that
$$
\lim_{x\rightarrow \bar{x}}\frac{f(x)-f(\bar{x})-\nabla f(\bar{x})(x-\bar{x})}{\|x-\bar{x}\|}=0.
$$
$f$ is said to be strictly differentiable at $\bar{x}$ if $f$ is Frechet differentiable
at $\bar{x}$ and 
$$
\lim_{\substack{x\rightarrow \bar{x}\\ u\rightarrow \bar{x}}}\frac{f(x)-f(u)-\nabla f(\bar{x})(x-u)}{\|x-u\|}=0.
$$


It is known that if $f$ is continuously Frechet differentiable in a neighborhooh of $\bar{x}$
 then $f$ is strictly differentiable at this point but not vice versa.

I would like to find a mapping $f$ such that $f$ is strictly differentiable at $\bar{x}$
but it is not differentiable at points near $\bar{x}$.
What you call strict differentiability is what Schechter calls strongly differentiable. See chapter 25 of his book. 

Section 25.23 proves that


  If $f$ is Frechet differentiable on an open set, then $f$ is continuously differentiable if and only if $f$ is strongly differentiable on the same set. 


Now to answer your question:

Let $f:\mathbb{R}\to\mathbb{R}$ be given by (below we have $j\in\mathbb{Z}$):
$$ f(t) = \begin{cases} 2^{2j} + (|t| - 2^{j}) ( 2^{j+2} - 2^j) &amp; |t|\in [2^j, 2^{j+1}) \\ 0 &amp; t = 0 \end{cases} $$
Our function $f(t)$ is piecewise linear and continuous, as can be easily checked. It is not differentiable at any $t = 2^j$ since the slope at $(2^j)_+$ and $(2^j)_-$ are different. 

One can, however, easily check that $f$ is strongly differentiable at 0 with derivative 0. We see easily that for $x,y$
$$ \frac{|f(x) - f(y)|}{|x-y|} \leq \frac{|f(|x|) - f(|y|)}{||x| - |y||} $$
so we can assume WLOG $x,y$ has the same sign. Then we have that using the convexity of the construction (the slope increases), that $f(x) - f(y)$ (assuming $x &gt; y$) is at most $3x(x-y)$. And hence as $x,y \to 0$ the limit exists and is 0.
$f: [-1,1] \rightarrow \mathbb R , \bar x=0$, defined by
$f(x):=\begin{cases} x^2 &amp; x=1/k , k\in \mathbb N \\ 0 &amp; x=0\\ \text{linear} &amp; \text{otherwise} \end{cases}$