Given the Cauchy problem $y' = \frac{\sin y + y}{(x^2 + 1)^{-2}e^{\left(\frac{x^3}{3} + x\right)}}$, $y(x_0) = y_0$, where $x_0 \in \mathbb{R}^+$$\cup$ {$0$}.
a) Does $f(x,y)$ in the vicinity of $(x_0,y_0)$ satisfy the local existence theorem?
b) Using the global existence theorem, verify if the solution to the equation is unique on $\mathbb{R}^+$$\cup$ {$0$}.
Attempt: The function $f$ is obviously continuous since $x^2+1 \geq 1 > 0$ and $e^{(x^3/3+x)} > 0$. Then I looked at what $|f(x,y_2)-f(x,y_1)|$ is equal to:
\begin{split}&\left|\frac{\sin y_2 + y_2}{(x^2+1)^{-2}e^{(x^3/3+x)}} - \frac{\sin y_1 + y_1}{(x^2+1)^{-2}e^{(x^3/3+x)}}\right| \\&= \left|\frac{\sin y_2 + y_2 - \sin y_1 - y_1}{(x^2+1)^{-2}e^{(x^3/3+x)}}\right|\end{split}
But now I don't know how to estimate this to be $<M|y_2 - y_1|$. Any idea how to do it? Thanks in advance!
Added attempt:
After the comments below, I have now considered that $|f(x,y_2)-f(x,y_1)|=\frac{1}{(x^2+1)(e^{x^3/3+x})}|(\sin y_2+y_2)-(\sin y_1+y_1)|$, which, according to the Lagrange theorem, is equal to $\frac{1}{(x^2+1)(e^{x^3/3+x})}|cos \theta+1||y_2-y_1|\leq \frac{1}{(x^2+1)(e^{x^3/3+x})}2|y_2-y_1|$. Is this okay now and how should I show that $2\frac{1}{(x^2+1)(e^{x^3/3+x})}<M$ for some $M>0$?