No, the diagonal being zero does not mean the matrix must be non-invertible. Consider $\begin{pmatrix} 0 & 1 \\ -1 & 0 \\ \end{pmatrix}$. This matrix is skew-symmetric with determinant $1$.
Edit: as a brilliant comment pointed out, it is the case that if the matrix is of odd order, then skew-symmetric will imply singular. This is because if $A$ is an $n \times n$ skew-symmetric we have $\det(A)=\det(A^T)=det(-A)=(-1)^n\det(A)$. Hence in the instance when $n$ is odd, $\det(A)=-\det(A)$; over $\mathbb{R}$ this implies $\det(A)=0$.
A construction can be found in lemma 5.2.2, pp.36-37 of Olga Ruff's master thesis The Jordan canonical forms of complex orthogonal and skew-symmetric matrices: characterization and examples.
To summarise, let $z=\frac{1-i}{2}$. Since $\pmatrix{z&\overline{z}\\ \overline{z}&z}^2=\pmatrix{0&1\\ 1&0}$, if we set $X$ to the $(2n+1)\times(2n+1)$ matrix
$$
\pmatrix{
z&&&&&&&&&&\overline{z}\\
&iz&&&&&&&&i\overline{z}\\
&&z&&&&&&\overline{z}\\
&&&iz&&&&i\overline{z}\\
&&&&\ddots&&\unicode{x22F0}\\
&&&&&\sqrt{(-1)^n}\\
&&&&\unicode{x22F0}&&\ddots\\
&&&i\overline{z}&&&&iz\\
&&\overline{z}&&&&&&z\\
&i\overline{z}&&&&&&&&iz\\
\overline{z}&&&&&&&&&&z},
$$
then
\begin{aligned}
X^2&=\operatorname{antidiag}(1,-1,1,-1,\ldots,1)=DR=RD,\text{ where}\\
D&=\operatorname{diag}(1,-1,1,-1,\ldots,1),\\
R&=\operatorname{antidiag}(1,1,\ldots,1).
\end{aligned}
Let $J=J_{2n+1}(0)$. Since $X$ is symmetric and $X^4=I$, we have
$$
(XJX^{-1})^T=X(X^2J^TX^2)X^{-1}
=XDRJ^TRDX^{-1}=XDJDX^{-1}=-XJX^{-1},
$$
i.e. $K=XJX^{-1}$ is skew-symmetric and similar to $J$.
We can prove by a parity argument that nilpotent Jordan blocks of even sizes are not similar to any complex skew-symmetric matrices. First, we need the following result of Horn and Merino (2009) (which is also part of lemma 5.1.2 in Olga Ruff's thesis).
Lemma. A complex square matrix $A$ is similar to a complex skew-symmetric matrix $K$ only if $SA$ is skew-symmetric for some complex symmetric matrix $S$.
Proof. If $A=P^{-1}KP$ where $K^T=-K$, then $A^T=-(P^TP)A(P^TP)^{-1}$. Hence $P^TPA$ is skew-symmetric. $\square$
Now suppose an $m\times m$ nilpotent Jordan block $J=J_m(0)$ is similar to a skew-symmetric matrix. By the above lemma, $SJ$ is skew-symmetric for some non-singular symmetric matrix $S$. Note that the first column of $SJ$ is zero. Therefore
$$
S_{1j}=(SJ)_{1,j+1}=-(SJ)_{j+1,1}=0 \textrm{ for all } j<m.\tag{1}
$$
Moreover, by the symmetry of $S$ and skew-symmetry of $SJ$,
$$
S_{ij}=S_{ji}=(SJ)_{j,i+1}=-(SJ)_{i+1,j}=-S_{i+1,j-1}.\tag{2}
$$
Equality $(1)$ means that all entries on the first row of $S$ except the rightmost one are zero. Equality $(2)$ means that if we travel down an anti-diagonal of $S$, the entries are basically constant but they have alternating signs. It follows from $(1)$ and $(2)$ that all entries of $S$ above the main anti-diagonal are zero and the main anti-diagonal of $S$ is $\left(s,-s,s,-s,\ldots,(-1)^{m-1}s\right)$ for some $s$. As $S$ is non-singular, $s$ must be nonzero. Yet, as $S$ is symmetric, the first and the last entries on the anti-diagonal must be equal. Hence $s=(-1)^{m-1}s$ and $m$ is odd.
Best Answer
Use the property $(AB)^t=B^tA^t$ to compute the transpose of each matrix and the fact that $A$ is symmetric and $B$ is skew-symmetric. For example, $(ABA)^t=A^tB^tA^t=A(-B)A=-(ABA)$. Then $ABA$ is skew-symmetric (and not symmetric in general).