Prove that limit doesn't exists at all

Because there exists a subsequence converging to $0$:

if you pick $\{x_k\}$ to be such that $x_k=\frac{1}{\pi + 2\pi k}$, then $$f^\prime (x_k) = -(\pi + 2\pi k)^2 \Big(1+cos(\pi + 2\pi k)\Big) = 0,$$ since $$cos(\pi + 2\pi k)=-1 \mbox{, } \forall k;$$ Also the sequence of $x_k$ converges to $0$ as you can observe letting $k\to +\infty$.

With $z:=\dfrac1x$, rewrite as

$$\lim_{z\to\infty}-z^2(1+\cos z).$$

Then it is clear that the function alternates between $0$ and $-2z^2$.

Scrap the whole proof by contradiction thing, it's not necessary.

Hint: Find the values of $x$ such that $\frac{1}{x} = 2k\pi$, and $\frac{1}{x} = (2k+1)\pi,$ where $k \in \mathbb{Z}$.

Edit: In fact, as Calvin Khor pointed out in the comments to this answer, you only need one of the above two sequences to show that $\lim_{x \to0^+}f'(x) \ne -\infty$, which was the original question.

Further edit: I think OP doesn't properly understand what is meant by $lim_{x \to c^+}g(x) = -\infty$. I also needed a sanity check, which is why I posted this question.

If $g(x)$ is a real function, then a definition of $\lim_{x \to c^+}g(x) = -\infty$ is:

for each $\gamma \in \mathbb{R}, \exists \delta > 0 $ such that $x \in (c,c+\delta) \implies g(x) < \gamma$.

So an equivalent definition of $\ \neg\left(\lim_{x \to c^+}g(x) = -\infty\ \right) $ is:

there exists $\beta\in\mathbb{R}$ such that for every $\delta>0$, there exists $x'\in(c,c+\delta)$ with $g(x')\ge \beta$

If no such $\beta$ exists, then $\ \lim_{x \to c^+}g(x) = -\infty\ .$

Note that $g$ need not be continuous at $c$ in either of my definitions.