Prove if $ x^{11}\equiv 21$ (mod $59$) then $x^{9}\equiv 35$ (mod $59$) Prove if $ x^{11}\equiv 21$ (mod $59$) then $x^{9}\equiv 35$ (mod $59$).
I know by Fermatt's Little Theorem, $x^{58} \equiv1$ (mod 59), and I know that $x^{58} = (x^{11})^5(x^3)$, but I dont know where to go from here. What do I do next?

As you figured out
$$(x^{11})^5 \equiv (21)^5 \equiv 3 \pmod{59} \Rightarrow (x^{11})^5x^3 \equiv 3x^3 \pmod{59} \Rightarrow \\
1 \equiv 3x^3 \pmod{59} \Rightarrow 1 \equiv 3^3x^9 \pmod{59}\Rightarrow \\
35 \equiv (35\cdot 27) \cdot x^9 \equiv x^9 \pmod{59}$$
since
$$35\cdot 27 \equiv 1 \pmod{59}$$
