$f(x)$ is a cubic function with leading coefficient of 1, $$ g(x) = f(x-1) \times \left|\lim_{ h\to 0} \frac{f(x+h)-f(x)}{h}\right| $$
$g(x)$ is not differentiable only at $x= -1$.
$g(1)=0$, $f'(0)>0$
All possible values of $f(1)$ ?
It is obvious that $g(x) = f(x-1) \left|f'(x)\right|$, and since $f(x-1)$ is a simple polynomial, non-differentiable point must occur at $\left|f'(x)\right|$. And it is a quadratic function, so I thought it could have $x=-1$ as one of its roots.
But applying the definition of derivative on $g(x)$ requires too much work because I had to set $f(x) = x^3+ ax^2+bx$. I don't have any clue on how to see $g(x)$ from different view. A little hint could be really helpful!
By using a hint from @StephenDonovan, $f'(x) = 3(x+1)(x-\alpha)$, and $\alpha \ne -1$ since if so, it becomes differentiable on every $x$, which is a contradiction.
Using $f'(0)>0$, we can get $\alpha < 0$. Using $g(1)=0=f(0)f'(1)$, $f(0)=0 (\because f'(1) \ne 0)$.
By integration, $f(x) = x^3 + \frac{3}{2}(1-\alpha)x^2 - 3\alpha x$.
And because $g(x)$ has to be differentiated on $x=\alpha$, $$\lim_{x \to \alpha} \frac{g(x)-g(\alpha)}{x-\alpha} \\ = \lim_{x \to \alpha} \frac{f(x-1)\times \left| f'(x) \right|}{x-\alpha} \\ \lim_{x \to \alpha} \frac{f(x-1)\times \left| 3(x+1)(x-\alpha) \right|}{x-\alpha}$$ must exist.
By taking $x \to \alpha ^\pm$, $f(\alpha - 1) = 0$. By substitution, we can get $\alpha = -2 \pm \sqrt 3$, and therefore getting all $f(1)$s.