565. $y=\frac{1}{3} x^{3}-2 x^{2}+3 x+1$.
Solution. $1^{0}$. Find the derivative $y^{\prime}=x^{2}-4 x+3$.

$2^{0}$. Set it to zero and solve the equation $x^{2}-4 x+3=0$. Its roots $x_{1}=1, x_{2}=3$ are critical points.

$3^{0}$. The derivative can be represented as a product of factors: $y^{\prime}=(x-1)(x-3)$. Investigate the critical point $x_{1}=1$, determining the sign of $y^{\prime}$ near this point to the left and right of it. Since $y_{x<1}^{\prime}<0$, $y_{x>1}^{\prime}>0$, the function reaches a minimum at $x_{1}=1$. Similarly, since $y_{x<3}^{\prime}>0$, $y_{x>3}^{\prime}<0$, the function reaches a maximum at $x_{2}=3$.

$4^{0}$. Find $\left.y\right|_{x=1}=\frac{7}{3},\left.y\right|_{x=3}=1$.

The graph of the function is shown in Fig. 122.