To solve the functional equation for all functions $f : \mathbb{Q}^+ \to \mathbb{Q}^+$ such that for all $x, y \in \mathbb{Q}^+$,

$$f(f(x)^2 y) = x^3 f(xy),$$

we proceed with the following steps:

Step 1: Simplify the equation using a special substitution.

First, consider setting $y = 1$. The equation becomes:

$$f(f(x)^2) = x^3 f(x).$$

This relationship will help us understand how $f$ behaves when applied to inputs derived from $f(x)$.

Step 2: Making another strategic substitution.

Let us choose $x = 1$ and substitute it back into the original equation:

$$f(f(1)^2 y) = f(y).$$

This implies that for any positive rational number $y$, $f$ is periodic in respect to an argument of the form $f(1)^2 y$.

**Step 3: Inferring a potential form of the function $f$.**

Consider the function $f(x) = \frac{1}{x}$. Check if this satisfies the given functional equation:

Calculate $f(f(x)^2 y)$ with $f(x) = \frac{1}{x}$:

- $f(x)^2 = \frac{1}{x^2}$,
- $f(f(x)^2 y) = f\left(\frac{1}{x^2} y\right) = \frac{1}{\frac{1}{x^2} y} = x^2 \cdot \frac{1}{y} = \frac{x^2}{y}$.

Now, calculate $x^3 f(xy)$:

- $f(xy) = \frac{1}{xy}$,
- $x^3 f(xy) = x^3 \cdot \frac{1}{xy} = \frac{x^3}{xy} = \frac{x^2}{y}$.

The two expressions are equal, thus confirming that $f(x) = \frac{1}{x}$ is indeed a valid solution.

Step 4: Conclude the findings.

Based on the exploration, the only function satisfying the given functional equation is:

$$\boxed{f(x) = \frac{1}{x}}$$

for all $x \in \mathbb{Q}^+$.