The given differential equation is $(x + 3y^2) \frac{dy}{dx} = y$. To find the general solution, we first rearrange the terms to express it in the standard linear form. Since the equation is linear in terms of $x$, we write it in the form $\frac{dx}{dy} + P(y)x = Q(y)$.

Rearranging the equation, we have $\frac{dy}{dx} = \frac{y}{x + 3y^2}$. Inverting both sides gives $\frac{dx}{dy} = \frac{x + 3y^2}{y}$. Simplifying the right-hand side, we separate the terms to get $\frac{dx}{dy} = \frac{x}{y} + \frac{3y^2}{y}$. This leads to the standard linear form:

$$\frac{dx}{dy} - \frac{1}{y}x = 3y$$

Here, the coefficient of $x$ is $P(y) = -\frac{1}{y}$ and the term on the right is $Q(y) = 3y$.

Next, we determine the Integrating Factor (I.F.). The I.F. is given by $e^{\int P(y) \, dy}$. Substituting the value of $P(y)$, we calculate $\text{I.F.} = e^{\int -\frac{1}{y} \, dy}$. Solving the integration, we get $\int -\frac{1}{y} \, dy = -\ln y = \ln(y^{-1})$. Therefore, the Integrating Factor is $e^{\ln(y^{-1})} = y^{-1} = \frac{1}{y}$.

Using the solution formula for a linear differential equation, $x \cdot (\text{I.F.}) = \int Q(y) \cdot (\text{I.F.}) \, dy + C$, we substitute the values. Multiplying the standard form equation by the I.F. $\frac{1}{y}$, we get $\frac{x}{y} = \int 3y \cdot \frac{1}{y} \, dy + C$. Simplifying the integrand, the equation becomes $\frac{x}{y} = \int 3 \, dy + C$.

Finally, performing the integration on the right-hand side, we obtain $\frac{x}{y} = 3y + C$. Multiplying both sides by $y$ to isolate $x$, we arrive at the final expression. The general solution of the given differential equation is:

$$\mathbf{x = 3y^2 + Cy}$$