You can't use the thing you want to prove in the proof itself, so the above answers are missing some steps. Here is a more detailed and complete proof.

Given A is nonsingular and symmetric, show that $ A^{-1} = (A^{-1})^T $.

Since $A$ is nonsingular, $A^{-1}$ exists. Since $ I = I^T $ and $ AA^{-1} = I $,

$$ AA^{-1} = (AA^{-1})^T. $$

Since $ (AB)^T = B^TA^T $,

$$ AA^{-1} = (A^{-1})^TA^T. $$

Since $ AA^{-1} = A^{-1}A = I $, we rearrange the left side to obtain

$$ A^{-1}A = (A^{-1})^TA^T. $$

Since $A$ is symmetric, $ A = A^T $, and we can substitute this into the right side to obtain

$$ A^{-1}A = (A^{-1})^TA. $$

From here, we see that

$$ A^{-1}A(A^{-1}) = (A^{-1})^TA(A^{-1})$$
$$ A^{-1}I = (A^{-1})^TI $$
$$ A^{-1} = (A^{-1})^T, $$

thus proving the claim.