570. $y=x e^{x}$.
Solution. $1^{0}$. Find the derivative:

$$
y^{\prime}=\left(x e^{x}\right)^{\prime}=x^{\prime} e^{x}+\left(e^{x}\right)^{\prime} x=e^{x}+x e^{x}=e^{x}(1+x)
$$

$2^{0}$. Find the critical points: $e^{x}(1+x)=0, x=-1$.

$3^{\circ}$. Investigate the signs of the derivative to the left and right of the critical point:

$$
y^{\prime}(-2)=e^{-2}(1-2)=\frac{1}{e^{2}}(-1)<0, \quad y^{\prime}(0)=e^{0}(1+0)=1>0
$$

Therefore, at $x=-1$ the function has a minimum $y_{\min }=y(-1)=$ $=(-1) e^{-1}=-1 / e=-0.369$.

Let's construct a table:

| $x$ | $(\infty,-1)$ | -1 | $(-1, \infty)$ |
| :---: | :---: | :---: | :---: |
| $y^{\prime}$ | - | 0 | + |
| $y$ | decreases | $y_{\text {min }}=-1 / e$ | increases |