Let $f:[0,1]\to\mathbb{R}$ be such that $f(xy)=f(x)\,f(y)$ for all $x,y\in[0,1]$, and $f(0)\ne 0$. If $y=v(x)$ satisfies the differential equation $\dfrac{dy}{dx}=f(x)$ with $y(0)=1$, then $y\!\left(\dfrac{1}{4}\right)+y\!\left(\dfrac{3}{4}\right)$ is equal to: