The statement is not generally true. Take for example $f(x) = \cos(x) + 10$ with $L=1$, then:
$$
\tilde{f}(x) = \lvert x\rvert(\cos(x/\lvert x\rvert) + 10) = \lvert x\rvert(\cos(1) + 10) ,
$$
which has smallest Lipschitz constant $\tilde{L} = \cos(1) + 10 > 3$.
However, let $x,y\in\mathbb{R}^n\setminus\{0\}, x\neq y,$ such that $\lVert x\rVert = \lVert y\rVert = a$, then
$$
\lvert\tilde{f}(x) - \tilde{f}(y)\rvert=a\lvert f(x/a) - f(y/a)\rvert\leq aL\lVert x/a - y/a\rVert = L\lVert x-y\rVert.
$$
Now, let $x,y\in\mathbb{R}^n\setminus\{0\}, x\neq y,$ be arbitrary and assume w.l.o.g. that $a = \lVert x\rVert > \lVert y \rVert$, and let $z = ay/\lVert y \rVert$, $t=\lVert y\rVert/a, 0 < t < 1$, then
$$
\lvert\tilde{f}(z) - \tilde{f}(y)\rvert = \lvert af(z/a) - taf(z/a)\rvert = a\lvert f(z/a)\rvert\lvert t-1\rvert \leq a(1-t)(\lvert f(z/a) - f(0)\rvert + \lvert f(0)\rvert)\leq a(1-t)(L + \lvert f(0)\rvert).
$$
Therefore
$$
\lvert\tilde{f}(x) - \tilde{f}(y)\rvert \leq \lvert\tilde{f}(x) - \tilde{f}(z)\rvert + \lvert\tilde{f}(z) - \tilde{f}(y)\rvert \leq L\lVert x-z\rVert + a(1-t)(L + \lvert f(0)\rvert)\leq\\ L\lVert x-y\rVert + L\lVert y-z\rVert + a(1-t)(L + \lvert f(0)\rvert) = L\lVert x-y\rVert + taL\lvert 1/t - 1 \rvert + a(1-t)(L + \lvert f(0)\rvert)=\\
L\lVert x-y\rVert + aL(1-t) + a(1-t)(L + \lvert f(0)\rvert) = L\lVert x-y\rVert + a(1-t)(2L + \lvert f(0)\rvert).
$$
Now,
$$
\lVert x \rVert = a \leq \lVert x-y\rVert + at \implies \frac{1}{ \lVert x-y\rVert}\leq \frac{1}{a(1-t)},
$$
thus
$$
\frac{\lvert\tilde{f}(x) - \tilde{f}(y)\rvert}{\lVert x-y\rVert}\leq L + \frac{a(1-t)(2L + \lvert f(0)\rvert)}{\lVert x-y\rVert} \leq 3L + \lvert f(0)\rvert.
$$
Because the choice of $x$ and $y$ is arbitrary, we conclude that the Lipschitz constant of $\tilde{f}$ is no larger than $3L + \lvert f(0)\rvert$.
Your method works, because $|x| \leq 3$ on this domain, so making that replacement gives a Lipschitz constant. However you actually discarded the division by 2 for no real reason so you ended up with a suboptimal constant.
A faster way to do it is to use a bound on the partial derivative:
$$\left | \frac{\partial f}{\partial y} \right | =\left | x\cos(xy) \right | \leq |x| \leq 3.$$
The fact that this is sufficient is a consequence of the mean value theorem.
Note that my result is the same as what you would get from your method if you didn't discard the division by 2.
Best Answer
If $f$ is differentiable in its domain D than, if its derivative is bounded in D, you have that $L_f=\sup\{|f^{'}(x)| \ : \ x \in D \}$.
The function is easily differentiable in each of the three open intervals (when you talk about derivative, you have to work on an open set because you want to do limits) but is not overall differentiable (the conjuction points are the problems).
So, for solving your problem, you can calculate the three Lipschitz constants via derivative (it's quite simple because the derivative is constant in all the three intervals) and then take as $L_f$ the maximum of these three values. Finally, you can check manually if the Lipschitz constant holds also for the boundary points (so for a generic $x$ and each of the 4 boundary points and then for every possible combination of these 4 points, which are six).