Inequality

Since $x, y > 0$ , we can apply AM-GM inequality and RMS-AM inequality respectively as follows:

$\begin{aligned} \sqrt {xy} & \le \frac {x+y}2 = 1 \\ \implies x^2y^2 & \le 1 & \small {\color{#3D99F6}\text{Equality occurs when }x=y=1} \end{aligned}$

$\begin{aligned} \sqrt {\frac {x^2+y^2}2} & \le \frac {x+y}2 = 1 \\ \implies x^2+y^2 & \le 2 & \small {\color{#3D99F6}\text{Equality occurs when }x=y=1} \end{aligned}$

$\implies x^2y^2(x^2+y^2) \le \boxed{2}$ .

Using $AM-GM$ inequality, $\dfrac{x+y}{2}\geq\sqrt{xy}\Rightarrow 1\geq xy.$ Here in this case we take $xy=1$ since $1$ is its max. value.
${(x+y)}^2=x^2+y^2+2xy\Rightarrow x^2+y^2=2.$
Therefore,the maximum value of $x^2y^2(x^2+y^2)=1^2(2)=\boxed2.$