Here is my work.
$$ e^{\sin z} = e^{\sin(x+iy)} = e^{\sin x \cos(iy) + \cos x \sin(iy) } = e^{\sin x \cosh y + i\cos x \sinh y} $$
So, $u = e^{\sin x \cosh y}$ and $v = e^{\cos x \sinh y}$. Then,
$$ \frac{\partial u}{\partial x} = \cos x \, e^{\sin x \cosh y} \qquad\text{and}\qquad \frac{\partial v}{\partial y} = \cosh y \, e^{\cos x \sinh y}. $$
So, not analytic because $\frac{\partial u}{\partial x} \neq \frac{\partial v}{\partial y}$ thus not satisfying Cauchy Riemann equations.
Is it correct?
No, it is not correct. We have indeed$$e^{\sin(x+yi)}=e^{\sin(x)\cosh(y)+\cos(x)\sinh(y)i},$$but this, in turn, is equal to$$e^{\sin(x)\cosh(y)}\left(\cos\bigl(\cos(x)\sinh(y)\bigr)+\sin\bigl(\cos(x)\sinh(y)\bigr)i\right).$$So$$u(x,y)=e^{\sin(x)\cosh(y)}\cos\bigl(\cos(x)\sinh(y)\bigr)$$and$$v(x,y)=e^{\sin(x)\cosh(y)}\sin\bigl(\cos(x)\sinh(y)\bigr).$$Anyway, $\exp\circ\sin$ is analytic because it is the composition of two analytic functions.