Answer
See the proof below.
Work Step by Step
Since $A$ is a nonsingular symmetric matrix, then $A$ is invertible and $A^{T}=A$
Since $A$ is invertible, then $(AA^{-1})^{T}=I^{T}=I$ and
$(A^{-1})^{T}A^{T}=I$
Since $A^{T}=A$, then $(A^{-1})^{T}A^{T}=I$ implies that $(A^{-1})^{T}A=I$
Since $A$ is invertible, then $A$ has an inverse and then
$((A^{-1})^{T}A)A^{-1})=IA^{-1})$
This implies that $(A^{-1})^{T} I=(A^{-1})^{T}=A^{-1}$
Thus $(A^{-1})^{T}=A^{-1}$
This indicates that $A^{-1}$ is symmetric.
