Check my general solution to the differential equation? Given differential equation:
$$y' = \frac{(2xy^{3}+4x)}{(x^{2}y^{2}+y^{2})}$$
This is the general solution that I got for the above differential equation:
$$\frac{1}{3} \ln{\lvert y^3+2\rvert}=\ln{\lvert x^2+1\rvert}+\ln C$$
Please check if it is correct?

And for the solution satisfying y(1)=0, this is what I got:
$$(y^3+2)^{\frac{1}{3}}=(x^2+1)+C$$

For the sake of completion, I'll run through the solution of the differential equation. First, our original equation is the same as
$$\frac{dy}{dx} = \frac{(2x)(y^{3} + 2)}{(x^{2} + 1)(y^{2})}.$$
By separation of variables, we get
$$ \frac{y^{2}}{y^{3}+2} dy = \frac{2x}{x^{2}+1} dx.$$
Equivalently,
$$ \frac{1}{3} \frac{3y^{2}}{y^{3}+2} dy = \frac{2x}{x^{2} + 1} dx.$$
Now, we note that this
$$ \frac{1}{3} \frac{d}{dy} \ln \left| y^{3} + 2 \right| = \frac{d}{dx} \ln \left| x^{2} + 1 \right|,$$
so we integrate to get
$$ \frac{1}{3} \ln \left| y^{3} + 2 \right| = \ln \left| x^{2} + 1 \right| + C.$$
So that part of your answer is correct; I've just renamed your $``\ln C"$ to just $C$.

There is, however, a problem with your solution to the initial value problem. Most importantly, there should not be any undetermined constants! To solve the initial value problem, simply plug $x = 1$ and $y = 0$ into the general solution of the differential equation to figure out what $C$ is. That is,
$$\begin{aligned}
\frac{1}{3} \ln \left| (0)^{3} + 2 \right| &= \ln \left| (1)^{2} + 1 \right| + C \\
\frac{1}{3} \ln 2 &= \ln 2 + C \\
-\frac{2}{3} \ln 2 &= C.
\end{aligned}$$
Hence, the solution to your initial value problem is
$$\frac{1}{3} \ln \left| y^{3} + 2 \right| = \ln \left| x^{2} + 1 \right| - \frac{2}{3} \ln 2.$$
