Ta có: $3xyz=x^2+y^2+z^2\ge xy+yz+zx \Rightarrow \dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\le3$
Ta có:$P=\dfrac{x^2}{x^4+yz}+\dfrac{y^2}{y^4+xz}+\dfrac{z^2}{z^4+xy}$
$\le\dfrac{x^2}{2x^2\sqrt{yz}}+\dfrac{y^2}{2y^2\sqrt{xz}}+\dfrac{z^2}{2z^2\sqrt{xy}}$
$=\dfrac{1}{2\sqrt{yz}}+\dfrac{1}{2\sqrt{xz}}+\dfrac{1}{2\sqrt{xy}}$
$\le\dfrac{1}{4}\left(\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{1}{z}+\dfrac{1}{x}+\dfrac{1}{x}+\dfrac{1}{y}\right)\le\dfrac{3}{2}$.
Vậy $\max P=\dfrac{3}{2} \Leftrightarrow x=y=z=1$.