Consider the differential equation :
$$$\frac{dy}{dx} = \frac{y^3}{2(xy^2 - x^2)}$$$
Statement-1: The substitution $$z = y^2$$ transforms the above equation into a first order homogeneous differential equation.
Statement-2: The solution of this differential equation is $$y^2 e^{-y^2/x} = C$$.

We are given the differential equation:

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

Statement-1 claims that substituting $$ z = y^2 $$ transforms this into a first-order homogeneous differential equation. Statement-2 claims that the solution is $$ y^2 e^{-y^2/x} = C $$. We will verify both statements step by step.

First, for Statement-1, substitute $$ z = y^2 $$. Differentiate both sides with respect to $$ x $$:

$$\frac{dz}{dx} = 2y \frac{dy}{dx}$$

Substitute the given $$\frac{dy}{dx}$$ into this equation:

$$\frac{dz}{dx} = 2y \cdot \frac{y^3}{2(xy^2 - x^2)} = \frac{2y \cdot y^3}{2(xy^2 - x^2)} = \frac{y^4}{xy^2 - x^2}$$

Factor the denominator:

$$xy^2 - x^2 = x(y^2 - x)$$

So:

$$\frac{dz}{dx} = \frac{y^4}{x(y^2 - x)}$$

Since $$ z = y^2 $$, we have $$ y^4 = (y^2)^2 = z^2 $$ and $$ y^2 = z $$. Substitute these:

$$\frac{dz}{dx} = \frac{z^2}{x(z - x)}$$

This simplifies to:

$$\frac{dz}{dx} = \frac{z^2}{xz - x^2}$$

To check if this is homogeneous, express it in terms of $$ v = \frac{z}{x} $$. Substitute $$ z = vx $$ into the right-hand side:

$$\frac{z^2}{xz - x^2} = \frac{(vx)^2}{x(vx) - x^2} = \frac{v^2 x^2}{vx^2 - x^2} = \frac{v^2 x^2}{x^2(v - 1)} = \frac{v^2}{v - 1}$$

Thus:

$$\frac{dz}{dx} = \frac{v^2}{v - 1}$$

Since the right-hand side is a function of $$ v $$ only, the equation is homogeneous. Therefore, Statement-1 is true.

Next, for Statement-2, we solve the homogeneous equation $$\frac{dz}{dx} = \frac{v^2}{v - 1}$$ where $$ v = \frac{z}{x} $$. Use the substitution $$ z = vx $$, so:

$$\frac{dz}{dx} = v + x \frac{dv}{dx}$$

Set equal to the right-hand side:

$$v + x \frac{dv}{dx} = \frac{v^2}{v - 1}$$

Rearrange:

$$x \frac{dv}{dx} = \frac{v^2}{v - 1} - v = \frac{v^2 - v(v - 1)}{v - 1} = \frac{v^2 - (v^2 - v)}{v - 1} = \frac{v}{v - 1}$$

So:

$$x \frac{dv}{dx} = \frac{v}{v - 1}$$

Separate variables:

$$\frac{v - 1}{v} dv = \frac{1}{x} dx$$

Simplify:

$$\left(1 - \frac{1}{v}\right) dv = \frac{1}{x} dx$$

Integrate both sides:

$$\int \left(1 - \frac{1}{v}\right) dv = \int \frac{1}{x} dx$$ $$v - \ln|v| = \ln|x| + C_1$$

Where $$ C_1 $$ is a constant. Rearrange:

$$v - \ln|v| - \ln|x| = C_1$$ $$v - \ln|vx| = C_1$$

Exponentiate both sides to eliminate the logarithm. First, write:

$$v - \ln|vx| = C_1$$

Multiply both sides by -1:

$$-v + \ln|vx| = -C_1$$

Exponentiate:

$$e^{\ln|vx| - v} = e^{-C_1}$$ $$|vx| e^{-v} = e^{-C_1}$$

Let $$ k = e^{-C_1} $$, a positive constant. Assuming positive values for simplicity, drop the absolute value:

$$vx e^{-v} = k$$

Substitute $$ v = \frac{z}{x} $$:

$$\left(\frac{z}{x}\right) x e^{-z/x} = k$$ $$z e^{-z/x} = k$$

Recall $$ z = y^2 $$:

$$y^2 e^{-y^2 / x} = k$$

Let $$ C = k $$, a constant. Thus:

$$y^2 e^{-y^2 / x} = C$$

This matches Statement-2. Therefore, Statement-2 is also true.

Both statements are true. Hence, the correct answer is Option D.