Let $$f(y)=\begin{cases}\sqrt{|y|}\sin\left(\frac{1}{y}\right) &\text{if} \ y\neq 0 \\ 0 &\text{if} \ y = 0\end{cases}$$ Prove that $f(y)$ is not Lipschitz (around zero).
Proof. It is easy to check that $f$ is continuous at zero. Now, for $y>0$, $f$ is differentiable and $$f'= \frac{\sin\left(\frac{1}{y}\right)}{2\sqrt{y}}- \frac{\sqrt{y}\cos\left(\frac{1}{y}\right)}{y^2} \geq -\frac{1}{2\sqrt{y}}-\frac{\sqrt{y}}{y^2}.$$
Also, $$\lim_{x\to 0^+}g(y)=\lim_{x\to 0^+}\frac{-y(y+2)}{2y^{5/2}}$$ gives the indeterminate form $\frac{0}{0}$. Thus, using L'Hopital's rule, this limit is $-\infty$, so $f'$ is not bounded near zero, and therefore cannot be Lipschitz.
Is this proof correct?
Alternative proof sketch. Let $A\lesssim B$ mean that $A\leq CB$ for some unimportant constant $C>0$. Consider the points given by $$y_n = \frac1{π(2n+1/2) } ⇔ \frac1{y_n}= π(2n+1/2) $$ Then $\sin \frac1{y_n} = 1$ for every $n$, so
$$ f(y_n) = \sqrt{y_n} = \frac1{\sqrt{π(2n+1/2)} } \gtrsim \frac 1{\sqrt{n}}$$ If $f$ were Lipschitz at 0, this would mean that
$$\frac 1{\sqrt{n}} \lesssim f(y_n) = |f(y_n) - f(0)| \leq L|y_n - 0| = Ly_n \lesssim\frac 1{n} $$
Which is false for $n\gg1$.
(PS you can use these $y_n$s with your computed derivative and the comment above regarding unbounded derivatives to fix your proof)