Inspired by Aditya Kumar

$\begin{aligned} x^4+y^4+z^4 & = (\color{#3D99F6}{x^2+y^2+z^2})^2 -2(x^2y^2+y^2z^2 + z^2 x^2) \quad \quad \small \color{#3D99F6}{\text{Since } x^2+y^2+z^2 =1} \\ & = 1 - 2(\color{#3D99F6}{x^2y^2+y^2z^2 + z^2 x^2}) \quad \quad \quad \quad \quad \quad \quad \quad \small \color{#3D99F6}{\text{Note that }x^2y^2+y^2z^2 + z^2 x^2 \ge 0 } \\ \Rightarrow x^4+y^4+z^4 & \le \boxed{1} \end{aligned}$

Maximum value of $x^4+y^4+z^4$ is $1$ if and when $x^2y^2+y^2z^2 + z^2 x^2 = 0$ . We note that when $x=y=0$ and $z = 1$ , then $x^2+y^2+z^2 =1$ and $x^2y^2+y^2z^2 + z^2 x^2 = 0$ .

We have $x^2\leq 1$ so $x^4\leq x^2$ . Likewise, $y^4\leq y^2$ and $z^4\leq z^2$ so $x^4+y^4+z^4\leq x^2+y^2+z^2=\boxed{1}$ . The maximum is attained when $x=1$ and $y=z=0$ , for example.

This might not be a proper approach. But it helps me many times.

$let\quad f:\quad { x }^{ 4 }+y^{ 4 }+{ z }^{ 4 }\quad \\ For\quad max\quad value,\quad \quad \quad \quad \quad ({ x }^{ 2 }+y^{ 2 }+{ z }^{ 2 }=1)\\ \frac { \partial f }{ \partial x } =0\\ \Rightarrow x=0\\ Similarly\quad y=z=0.\\ But\quad this\quad is\quad minima.\quad Using\quad the\quad binding\quad provided,\\ any\quad 2\quad variables,\quad say\quad x=y=0\quad and\quad z=1.\\ \therefore \quad max\left\{ { x }^{ 4 }+y^{ 4 }+{ z }^{ 4 } \right\} =1$

Please correct me