$f(x)=\begin {cases}\sin x\ln x^2 & x\neq 0\\ 0 & x=0\end{cases}$
When I try to find the derivative on $x=0$ with the defintion I get:
$\displaystyle\lim_{h\to 0}\frac {f(h+0)-f(0)}{h-0}=\lim_{h\to 0}\frac {f(h)}{h}=\lim_{h\to 0}\frac {\sin h \ln h^2} h=\lim_{h\to 0}\ln h^2=-\infty$
But I see from the graph that the slope on $x=0$ is supposed to be finite:
Why does the derivative definition don't work here?
the function $f$ defined by $f(x) = \sin x \ln(x^2)$ is an odd function. $f$ is continuous at $x = 0$ and $$f(x) = 2x \ln x \text{ for } x \to 0+$$ the one sided derivative $$D_+f\rvert_0 = \lim_{x \to 0+} \frac{f(x)}{x} = \frac{2x\ln x }{x} = 2 \ln x = dne$$
the reason this is not obvious from your graph is the function $\ln x$ go to $-\infty$ much slower than say $1/x.$ if you zoom in around zero more and more you will see the slope getting bigger.
another way is to take the derivative of $x\ln x.$ you will find that $(x\ln x)' = 1+\ln x \to -\infty \text { as } x \to 0+ $
there is no difference in the limiting behaviors of $\sin x \ln(x^2) $ and $2x\ln x$ as $x \to 0+$. that is why i took the simpler one.