Proving single solution to an initial value problem

I have the following initial value problem $$ y' = \left|\frac{1}{1+x^2} + \sin{|x^2 +\arctan{y^2}|}\right|, \space y(x_0)=y_0 $$ for each $(x_0, y_0)\in \mathbb{R} \times \mathbb{R}$ I need to prove that there is a single solution defined on $\mathbb{R}$

I think I know the general idea for proving this, I need to show that for every box $J \times (-\infty, \infty)$ that includes $(x_0, y_0)$ the function $$f(x, y)=\left|\frac{1}{1+x^2} + \sin{|x^2 +\arctan{y^2}|}\right|$$ is Lipschitz continuous on the $y$ variable and thus it has a single solution in the box $J \times (-\infty, \infty)$ and if it has a solution in every box then it has a solution defined on $\mathbb{R}$.

But I have no idea how I can show Lipschitz continuity on $f(x, y)$. So far I've shown it by deriving $f(x, y)$ and showing that $|f_y(x,y)|\le L$ for each $(x, y)\in (\alpha, \beta) \times (\gamma, \delta)$ but $f(x,y)$ isn't derivable

Solution 1:

let $$g(x, y) = \sin{\left|x^2 + \arctan{(y^2)} \right|}=\sin{(x^2 + \arctan{(y^2)})}$$ This function is derivable, hence $$g_y(x, y)=\frac{2y}{y^4+1}\cos{(x^2 + \arctan{(y^2)})}$$ $g_y$ is continuous thus for each $J$ there is a constant $L_J$ so that $$\left| g(x, y_1) - g(x, y_2)\right| \le L_J\left| y_1-y_2\right|$$ therefore $$ \left| f(x, y_1) - f(x, y_2)\right| = \left| \left| \frac{1}{1+x^2} + g(x, y_1)\right| - \left| \frac{1}{1+x^2} + g(x, y_2)\right|\right| \le \left| \frac{1}{1+x^2} + g(x, y_1) - \frac{1}{1+x^2} - g(x, y_2)\right| = \left| g(x, y_1) - g(x, y_2) \right| \le L_J\left| y_1-y_2\right| $$ and $f(x, y)$ is Lipschitz continuous on the 𝑦 variable, and this is true for each box $J \times (-\infty, \infty)$ and therefore there is a single solution on $\mathbb{R}$ for each $(x_0, y_0)\in\mathbb{R}\times\mathbb{R}$