Step 1

Let

$g(x,y)=\mathrm{sin}|{x}^{2}+\mathrm{arctan}\left({y}^{2}\right)|=\mathrm{sin}({x}^{2}+\mathrm{arctan}\left({y}^{2}\right))$

This function is derivable, hence

$g}_{y}(x,y)=\frac{2y}{{y}^{4}+1}\mathrm{cos}\left\{({x}^{2}+\mathrm{arctan}\left\{\left({y}^{2}\right)\right\})\right\$

$g}_{y$ is continuous thus for each J there is a constant $L}_{J$ so that

$|g(x,{y}_{1})-g(x,{y}_{2})|\le {L}_{J}|{y}_{1}-{y}_{2}|$

therefore

$|f(x,{y}_{1})-f(x,{y}_{2})|$

$=||\frac{1}{1+{x}^{2}}+g(x,{y}_{1})|-|\frac{1}{1+{x}^{2}}+g(x,{y}_{2})\mid \mid$

$\le |\frac{1}{1+{x}^{2}}+g(x,{y}_{1})-\frac{1}{1+{x}^{2}}-g(x,{y}_{2})|$

$=|g(x,{y}_{1})-g(x,{y}_{2})|\le {L}_{J}|{y}_{1}-{y}_{2}|$

and $f(x,y)$ is Lipschitz continuous on the 𝑦 variable, and this is true for each box $J\times (-\mathrm{\infty},\mathrm{\infty})$ and therefore there is a single solution on $\mathbb{R}$ for each $({x}_{0},{y}_{0})\in \mathbb{R}\times \mathbb{R}$