Proving that, $|f'(x)-f'(y)|\le k|x-y| \implies (f'(x))^2< 2 kf(x) $

Fix $x \in \Bbb R$.

$f'$ is (Lipschitz) continuous and therefore integrable. For any $d \ge 0$ we have $$ 0 < f(x+d) = f(x) + \int_x^{x+d} f'(t) \, dt = f(x) + df'(x) + \int_x^{x+d}(f'(t) - f'(x)) \, dt \\ \le f(x) + df'(x) + \int_x^{x+d}k(t-x) \, dt \\ = f(x) + df'(x) + \frac 12 kd^2 $$ and the same estimate holds for $d < 0$, because then $$ \int_x^{x+d}(f'(t) - f'(x)) \, dt = -\int_{x+d}^x(f'(t) - f'(x)) \, dt \le -\int_{x+d}^x k(t-x) \, dt = \frac 12 k d^2 \, . $$

In particular for $d = -\frac{f'(x)}{k}$ we get $$ 0 < f(x) - \frac{f'(x)^2}{k} + \frac{f'(x)^2}{2k} = f(x) - \frac{f'(x)^2}{2k} $$ which is the desired conclusion.

This is reminiscent of Landau's inequality.

WLOG $k=1$ (consider $g=f/k$).

Suppose $f'(0)^2 \ge 2f(0)$. WLOG $f'(0) \le -(2f(0))^{1/2}$ (if otoh $f'(0)$ is positive consider $g(t)=f(-t)$.) Then for every $t>0$ we have $$f'(t)\le t-(2f(0))^{1/2}.$$Integrating, this shows that $$f(x)\le f(0)+x^2/2-2x(2f(0))^{1/2}$$for every $x>0$. If $x=(8f(0))^{1/2}$ it follows that $f(x)<0$.

So $f'(0)^2<2f(0)$. The general case follows ((let $g(t)=f(x+t)$.)