2yy' +3 = y^2 + 3x with initial value y(0) = 8
One way to start this is to note that $\displaystyle \left ( y^2 \right ) ' = 2y~y'$, so your equation becomes:
$\displaystyle \left ( y^2 \right )' + 3 = y^2 + 3x$
Now, $\displaystyle y^2$ is just a function, let's call it $\displaystyle z = y^2$, turning your equation into
$\displaystyle z' + 3 = z + 3x$
Can you finish from here?
-Dan