If $f\left(x\right)=\left\{\begin{array}{cc}{e}^{2x^3+x}& x>0\\ ax+b& x\le 0\end{array}\right.$ is differentiable at $x=0,$ then

$$\begin{gathered} f^{\text{'}}\left(0^{-}\right)=\underset{h\to 0^{+}}{\lim }\frac{f\left(0-h\right)-f\left(0\right)}{-h} \\ \Rightarrow \underset{h\to 0^{+}}{\lim }\frac{\left(-ah+b\right)-\left(b\right)}{-h}=a \end{gathered}$$
$$\begin{gathered} f^{\text{'}}\left(0^{+}\right)=\underset{h\to 0^{+}}{\lim }\frac{f\left(0+h\right)-f\left(0\right)}{h} \\ \Rightarrow \underset{h\to 0^{+}}{\lim }\frac{e^{2h^3+h}-b}{h}\dots \dots \left(\text{i}\right) \end{gathered}$$
For this limit to exist, it must be of $\frac{0}{0}$ form.
$\therefore e^{2\left(0\right)+0}-b=0\Rightarrow b=1\mathrm{}\mathrm{}\dots \dots ..\left(\mathrm{i}\mathrm{i}\right)$
From $\left(\text{i}\right)$

$$\begin{gathered} f^{\text{'}}\left(0^{+}\right)=\underset{h\to 0^{+}}{\lim }\frac{e^{2h^3+h}-1}{h} \\ \Rightarrow \underset{h\to 0^{+}}{\lim }\frac{e^{2h^3+h}-1}{\left(2h^3+h\right)}\times \left(\frac{2h^3+h}{h}\right) \end{gathered}$$
$\Rightarrow 1\times 1=1$
$\because f^{\text{'}}\left(0^{-}\right)=f^{\text{'}}\left(0^{+}\right)$
$\Rightarrow a=1$