Step 1: Set up the layered function.
Let $F(y)=\int_0^{\log y} e^{-1/\sqrt2}(1-t^2)\,dt$, so that $f(x)=\int_0^{e^{\sin x}}F(y)\,dy$.
Step 2: Differentiate the outer integral.
By the fundamental theorem, $f'(x)=F\big(e^{\sin x}\big)\cdot \frac{d}{dx}e^{\sin x}$.
Step 3: Outer derivative.
\[ \frac{d}{dx}e^{\sin x}=e^{\sin x}\cos x \]
Step 4: Plug in $x=\frac{\pi}{4}$.
Here $\sin\frac{\pi}{4}=\cos\frac{\pi}{4}=\frac{1}{\sqrt2}$ and $\log(e^{\sin x})=\frac{1}{\sqrt2}$.
Step 5: Evaluate $F$.
\[ F=e^{-1/\sqrt2}\Big[t-\tfrac{t^3}{3}\Big]_0^{1/\sqrt2}=e^{-1/\sqrt2}\cdot\frac{5}{6\sqrt2} \]
Step 6: Combine everything.
\[ f'\!\Big(\frac{\pi}{4}\Big)=e^{-1/\sqrt2}\cdot\frac{5}{6\sqrt2}\cdot e^{1/\sqrt2}\cdot\frac{1}{\sqrt2}=\frac{5}{12}\approx 0.417 \]
Rounded to one decimal, this is $0.4$.
\[ \boxed{0.4} \]