First I apologize that the format is bad, because it's my first time typing math equations using this forum, so I am not sure how it works.
Typing is always better than scanning. Please try to modify post to see how I did it. Observe special meaning of \$ ...\$ for inline math and \$\$...\$\$ for display math (occupying its own lines) in the source. -- V.I.Second, wherever it says Q(x,y), it is not Q actually, it's that Greek letter phi which I don't know how to type.
OK, I will change $Q$ to $\Phi$ but I am not sure why you don't like $Q$. 
So here we go.
Solution: Rewrite
function equation as
$$( 9x^2 + y - 1 ) + ( x - 4y )y' = 0$$
So we have $M_y(x,y) = 1 = N_x(x,y)$, so this
function equation is exact.
$$
\left\{\begin{aligned}
&\Phi_x(x,y) = 9x^2 + y - 1,\\
&\Phi_y(x,y) = x - 4y.
\end{aligned}\right.
$$
Above system was more complicated to type.By integrating $\Phi_x(x,y)$, we have $\Phi (x,y) = 3x^3 + xy - x + h(y)$
with arbitrary $h(y)$.
By differentiating
derived $\Phi (x,y)$ with respect to $y$ we have $\Phi_y(x,y) = x + h'(y)$
Note that $ \Phi_y(x,y) = x + h'(y) = x - 4y$.
Therefore we have $h'(y) = -4y$, and $h(y) = -2y^2$.
So
$$
\Phi (x,y) = 3x^3 + xy - x - 2y^2 = c.
$$
Now we plug $y(1) = 0$ into this equation, we have $3(1^3) + (1)(0) - 1 - 2(0^2) = c\implies c = 2$.
So the solution is
$$
3x^3 + xy - x - 2y^2 = 2.
$$
Good Job -- V.I.
