Let $a_1,a_2\dots,a_n$ are the roots of $f(x)$.For $x=a_i$ for some $i$ the inequality is obvious.Otherwise,let $h(x)=\frac{f'(x)}{f(x)}$, $h$ is continuous and differentiable in every $x\ne a_i$. Observe that,$$h'(x)=\frac{f(x)f''(x)-f'(x)^2}{f(x)^2}$$Hence it is enough to show, $h(x)$ is a non-increasing function. $f$ can be written as $$f(x)=b(x-a_1)(x-a_2)\dots(x-a_n)$$. Hence,$$h(x)=\sum_{i=1}^n\frac{1}{(x-a_i)}$$But observe that ,$$m\ge n\implies\frac{1}{(m-a_i)}\le \frac{1}{(n-a_i)}$$. It implies $h$ is a non-increasing function.We are done $\blacksquare$

An alternative, purely inequalities, solution. For $n:={\rm deg}(f) = 1$, since $f''=0$, we have the inequality trivially. Suppose $n\ge 2$, set $\textstyle f(x) = \prod_{1\le i\le n}(x-a_i)$, where $a_i$, $1\le i\le n$ are the roots of $f$. Then notice that \[ f' = f(x) \sum_{1\le i\le n}\frac{1}{x-a_i} \qquad\text{and}\qquad f'' = 2f(x) \sum_{1\le i<j\le n}\frac{1}{(x-a_i)(x-a_j)}. \]Now, fix $x\in\mathbb{R}$; and set $b_i:=(x-a_i)^{-1}$. In terms of $b_i$'s it suffices to show \[ 2f(x)^2 \sum_{1\le i<j\le n}b_i b_j \le f(x)^2 \left(\sum_{1\le i\le n}b_i\right)^2 \Leftrightarrow 2\sum_{1\le i<j\le n}b_ib_j \le \left(\sum_{1\le i\le n}b_i\right)^2. \]This is clear.