## Calculus 8th Edition

$\bf True$.
$\bf True$. For this equation to be separable we would have to able to write $3y-2x+6xy-1$ from the equation $y'=3y-2x+6xy-1$ in the form of $f(x)g(y)$ where $f(x)$ is function in $x$ only and $g(y)$ is a function of $y$ only, which can be written as: $$y=3y-2x+6xy-1$$$$y=6xy+3y-2x-1$$$$y=(3y-1)(2x+1)$$ Hence, we can write $y'=3y-2x+6xy-1$ in the form of $f(x)g(y)$ where $f(x)=2x+1$ and $g(y)=3y-1$