Given differential equation

$e^{\sin y}\cos y\,\frac{dy}{dx}+e^{\sin y}\cos x=\cos x$

with initial condition

$y(0)=0.$

Recognize a derivative

Since

$\frac{d}{dx}\big(e^{\sin y}\big)=e^{\sin y}\cos y\,\frac{dy}{dx},$

the differential equation becomes

$\frac{d}{dx}\big(e^{\sin y}\big)+e^{\sin y}\cos x=\cos x.$

Let

$u=e^{\sin y}.$

Then the equation is

$\frac{du}{dx}+u\cos x=\cos x.$

Solve the linear differential equation

Rewrite:

$\frac{du}{dx}+(\,\cos x\,)u=\cos x.$

The integrating factor is

$IF=e^{\int \cos x\,dx}=e^{\sin x}.$

Multiplying throughout by $e^{\sin x}$:

$e^{\sin x}\frac{du}{dx}+u e^{\sin x}\cos x= e^{\sin x}\cos x.$

Hence,

$\frac{d}{dx}\big(u e^{\sin x}\big)=e^{\sin x}\cos x.$

Integrating,

$u e^{\sin x}=\int e^{\sin x}\cos x\,dx = e^{\sin x}+C.$

So,

$u=1+Ce^{-\sin x}.$

Use the initial condition

Since $y(0)=0$,

$u(0)=e^{\sin 0}=1.$

Also from the solution,

$u(0)=1+Ce^{-\sin 0}=1+C.$

Therefore,

$1+C=1 \implies C=0.$

So,

$u=1.$

That is,

$e^{\sin y}=1.$

Hence,

$\sin y=0.$

Determine the actual solution using the initial condition

Since $y(0)=0$, the solution consistent with continuity is

$y(x)=0 \quad \text{for all } x.$

Therefore,

$y\left(\frac{\pi}{6}\right)=0, \qquad y\left(\frac{\pi}{3}\right)=0, \qquad y\left(\frac{\pi}{4}\right)=0.$

Evaluate the required expression

$1+y\left(\frac{\pi}{6}\right)+\frac{\sqrt3}{2}y\left(\frac{\pi}{3}\right)+\frac{1}{\sqrt2}y\left(\frac{\pi}{4}\right)$

$=1+0+\frac{\sqrt3}{2}\cdot 0+\frac{1}{\sqrt2}\cdot 0=1.$

Final answer

$\boxed{1}$

Comparison with stored correct answer

Stored correct answer = $1$.

My derived answer also equals $1$, so they agree.