Given: The function $x^2 - y^2 = c (x^2 + y^2)^2$ is to be verified as the general solution of the differential equation $(x^3 - 3xy^2) dx = (y^3 - 3x^2y) dy$.

Step 1: Differentiate the given solution implicitly with respect to $x$.

Differentiating both sides of $x^2 - y^2 = c (x^2 + y^2)^2$:

$$2x - 2y \frac{dy}{dx} = c \cdot 2(x^2 + y^2) \cdot \left(2x + 2y \frac{dy}{dx}\right)$$

Step 2: Simplify the differentiated equation.

$$x - y \frac{dy}{dx} = 2c (x^2 + y^2) \left(x + y \frac{dy}{dx}\right)$$

From the given solution, $c = \frac{x^2 - y^2}{(x^2 + y^2)^2}$

Substituting this value:

$$x - y \frac{dy}{dx} = \frac{2(x^2 - y^2)}{x^2 + y^2} \left(x + y \frac{dy}{dx}\right)$$

Step 3: Multiply both sides by $(x^2 + y^2)$ and rearrange.

$$(x^2 + y^2)\left(x - y \frac{dy}{dx}\right) = 2(x^2 - y^2)\left(x + y \frac{dy}{dx}\right)$$

Expanding and collecting terms with $\frac{dy}{dx}$:

$$x(x^2 + y^2) - y(x^2 + y^2) \frac{dy}{dx} = 2x(x^2 - y^2) + 2y(x^2 - y^2) \frac{dy}{dx}$$

Step 4: Collect all $\frac{dy}{dx}$ terms on one side.

$$-y(3x^2 - y^2) \frac{dy}{dx} = x^3 - 3xy^2$$

This gives:

$$\frac{dy}{dx} = \frac{3xy^2 - x^3}{3x^2y - y^3}$$

Step 5: Rewrite the given differential equation.

From $(x^3 - 3xy^2) dx = (y^3 - 3x^2y) dy$:

$$\frac{dy}{dx} = \frac{x^3 - 3xy^2}{y^3 - 3x^2y} = \frac{3xy^2 - x^3}{3x^2y - y^3}$$

Conclusion: Both expressions for $\frac{dy}{dx}$ are identical. Hence, $x^2 - y^2 = c (x^2 + y^2)^2$ is verified as the general solution of the given differential equation.

∴ $x^2 - y^2 = c (x^2 + y^2)^2$ is the general solution of the given differential equation.