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To solve this problem, we need to find the particular solution to the differential equation given by:

\[
y' = \frac{3x^2}{y^2}
\]

which passes through the point \((0, 2)\).

### Step 1: Separate Variables
The differential equation can be rewritten to separate the variables \(x\) and \(y\):

\[
y^2 \, dy = 3x^2 \, dx
\]

### Step 2: Integrate Both Sides
Integrate both sides of the equation:

\[
\int y^2 \, dy = \int 3x^2 \, dx
\]

The integrals are:

\[
\frac{y^3}{3} = x^3 + C
\]

Multiply both sides by 3 to simplify:

\[
y^3 = 3x^3 + C
\]

### Step 3: Solve for the Constant \(C\)
We know that the curve passes through the point \((0, 2)\). Substitute \(x = 0\) and \(y = 2\) into the equation to find \(C\):

\[
2^3 = 3(0)^3 + C
\]

\[
8 = C
\]

Thus, the equation of the curve is:

\[
y^3 = 3x^3 + 8
\]

### Step 4: Solve for \(y\)
Taking the cube root of both sides gives:

\[
y = \sqrt[3]{3x^3 + 8}
\]

### Conclusion
Comparing this with the given options, we see that the correct answer is:

\[
\textbf{B. } y = \sqrt[3]{3x^3 + 8}
\]

Would you like more details or have any questions about the process? Here are some related questions that might interest you:

1. How do you solve a differential equation using the method of separation of variables?
2. What is the importance of the constant of integration in solving differential equations?
3. How would you check if a given function is a solution to a differential equation?
4. What is an initial condition, and how does it affect the solution of a differential equation?
5. What are the steps to solve a non-linear differential equation?

**Tip:** When dealing with differential equations, always make sure to apply the initial condition after integrating to find the particular solution.

¿Cuál es la curva que resuelve la ecuación y' = 3x^2 / y^2 y pasa por el punto (0, 2)?

Find the solution to the differential equation y' = 3x^2 / y^2 that passes through the point (0, 2). Learn how to solve using separation of variables and initial conditions, with a step-by-step explanation.

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