Let $a, b, c, d, e \in \mathbb{R}^+$ and $f:\{(x, y) \in (\mathbb{R}^+)^2|c-dx-ey > 0\}\to \mathbb{R}^+$ be given by $f(x, y) = (ax)(by)(c- dx- ey)$. 
Find the maximum value of $f$.
1. We start with the function \( f(x, y) = (ax)(by)(c - dx - ey) \) where \( a, b, c, d, e \in \mathbb{R}^+ \) and \( (x, y) \in (\mathbb{R}^+)^2 \) such that \( c - dx - ey > 0 \).

2. To find the maximum value of \( f(x, y) \), we use the Arithmetic Mean-Geometric Mean (AM-GM) inequality. The AM-GM inequality states that for any non-negative real numbers \( x_1, x_2, \ldots, x_n \),
   \[
   \frac{x_1 + x_2 + \cdots + x_n}{n} \geq \sqrt[n]{x_1 x_2 \cdots x_n}
   \]
   with equality if and only if \( x_1 = x_2 = \cdots = x_n \).

3. Applying the AM-GM inequality to the terms \( dx \), \( ey \), and \( c - dx - ey \), we get:
   \[
   \frac{dx + ey + (c - dx - ey)}{3} \geq \sqrt[3]{(dx)(ey)(c - dx - ey)}
   \]
   Simplifying the left-hand side, we have:
   \[
   \frac{c}{3} \geq \sqrt[3]{(dx)(ey)(c - dx - ey)}
   \]
   Cubing both sides, we obtain:
   \[
   \left( \frac{c}{3} \right)^3 \geq (dx)(ey)(c - dx - ey)
   \]

4. Therefore, we have:
   \[
   f(x, y) = (ax)(by)(c - dx - ey) \leq ab \left( \frac{c}{3} \right)^3 \frac{1}{de}
   \]
   Simplifying, we get:
   \[
   f(x, y) \leq \frac{a}{d} \frac{b}{e} \left( \frac{c}{3} \right)^3
   \]

5. Equality holds in the AM-GM inequality if and only if \( dx = ey = c - dx - ey \). Solving these equations, we get:
   \[
   dx = ey = \frac{c}{3}
   \]
   Therefore,
   \[
   x = \frac{c}{3d} \quad \text{and} \quad y = \frac{c}{3e}
   \]

6. We need to check if \( \left( \frac{c}{3d}, \frac{c}{3e} \right) \) satisfies the condition \( c - dx - ey > 0 \):
   \[
   c - d \left( \frac{c}{3d} \right) - e \left( \frac{c}{3e} \right) = c - \frac{c}{3} - \frac{c}{3} = c - \frac{2c}{3} = \frac{c}{3} > 0
   \]
   Thus, \( \left( \frac{c}{3d}, \frac{c}{3e} \right) \in \{(x, y) \in (\mathbb{R}^+)^2 | c - dx - ey > 0\} \).

7. Therefore, the maximum value of \( f(x, y) \) is:
   \[
   \frac{a}{d} \frac{b}{e} \left( \frac{c}{3} \right)^3
   \]

The final answer is \( \boxed{\frac{a}{d} \frac{b}{e} \left( \frac{c}{3} \right)^3} \).