We are given $y=e^x(\sin x-\cos x)$.
To find a point of inflection, we need $y''=0$ and the concavity must change sign.
First differentiate using the product rule:
$y'=e^x(\sin x-\cos x)+e^x(\cos x+\sin x)$
Simplify:
$y'=e^x(2\sin x)$
$y'=2e^x\sin x$
Differentiate again:
$y''=2e^x\sin x+2e^x\cos x$
$y''=2e^x(\sin x+\cos x)$
Now set $y''=0$:
$2e^x(\sin x+\cos x)=0$
Since $2e^x>0$ for all $x$, we only need
$\sin x+\cos x=0$
So,
$\tan x=-1$
In the interval $[0,\pi]$, this gives
$x=\frac{3\pi}{4}$
Now check whether the sign of $y''$ changes at $x=\frac{3\pi}{4}$.
Since $2e^x>0$, the sign of $y''$ depends only on $\sin x+\cos x$.
For $0<x<\frac{3\pi}{4}$, $\sin x+\cos x>0$, so $y''>0$ and the curve is concave up.
For $\frac{3\pi}{4}<x<\pi$, $\sin x+\cos x<0$, so $y''<0$ and the curve is concave down.
Therefore, concavity changes at $x=\frac{3\pi}{4}$.
Answer: B. $x=\frac{3\pi}{4}$
