Solve the differential equation. \[ \frac{d y}{d x}=9 x^{2} y^{2} \]

We begin with the differential equation \[ \frac{dy}{dx} = 9x^2 y^2. \] **Step 1. Separate the variables.** We rewrite the equation by dividing both sides by \( y^2 \) and multiplying by \( dx \): \[ \frac{1}{y^2} \, dy = 9x^2 \, dx. \] **Step 2. Integrate both sides.** Integrate the left side with respect to \( y \) and the right side with respect to \( x \): \[ \int \frac{1}{y^2} \, dy = \int 9x^2 \, dx. \] For the left side, we have \[ \int y^{-2} \, dy = -\frac{1}{y} + C_1. \] For the right side, we compute \[ \int 9x^2 \, dx = 9 \cdot \frac{x^3}{3} = 3x^3 + C_2. \] Combine the constants (let \( C = C_2 - C_1 \)): \[ -\frac{1}{y} = 3x^3 + C. \] **Step 3. Solve for \( y \).** Multiply the entire equation by \(-1\): \[ \frac{1}{y} = -3x^3 - C. \] Since \(-C\) is still an arbitrary constant, we can denote it by \( C \): \[ \frac{1}{y} = -3x^3 + C. \] Now, take the reciprocal of both sides to solve for \( y \): \[ y = \frac{1}{C - 3x^3}. \] Thus, the general solution of the differential equation is \[ \boxed{y = \frac{1}{C - 3x^3}}, \] where \( C \) is an arbitrary constant.