曲线的方程为 $y = x^2 e^{2x}$。

1. Identify the two factors and apply the product rule.

Let $u=x^2$ and $v=e^{2x}$.

$\frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}$

2. Differentiate each factor and simplify the expression.

$\frac{du}{dx}=2x, \qquad \frac{dv}{dx}=2e^{2x}$

So,
$\frac{dy}{dx}=x^2(2e^{2x})+e^{2x}(2x)$

$\frac{dy}{dx}=2x^2e^{2x}+2xe^{2x}$

$\frac{dy}{dx}=2xe^{2x}(x+1)$

1. Use the condition for a decreasing function.

For $y$ to decrease as $x$ increases,
$\frac{dy}{dx}<0.$
From part (i),
$2xe^{2x}(x+1)<0.$

2. Remove the always-positive factor from the inequality.

Since $e^{2x}>0$ for all real $x$, and $2>0$,
$2xe^{2x}(x+1)<0 \iff x(x+1)<0.$

3. Solve the sign inequality.

The critical values are
$x=0 \quad \text{or} \quad x+1=0,$
so
$x=0 \quad \text{or} \quad x=-1.$
The product $x(x+1)$ is negative between the two roots.
Therefore,
$-1<x<0.$