Analyze the symmetric property of positive definite matrices
Most authors define positive definite matrices as a subclass of symmetric matrices. This is not necessary, but then the usually equivalent definitions of positive definite matrices diverge.
Positive definite matrices by inner products
If you consider $\langle v, A v\rangle> 0,\forall v\in\Bbb R^n\setminus\{0\}$ as the defining property, then consider a matrix $A=S+T$, where $S$ is symmetric and positive definite (no definition problem because symmetric), and $T$ is a non-zero skew-symmetric matrix, i.e. $T^\top=-T$. Then $A$ is not symmetric, but positive definite because
$$\def\<{\langle} \def\>{\rangle}\< v, A v\>=\<v,(S+T)v\>=\underbrace{\<v,Sv\>}_{>\, 0}+\underbrace{\<v,Tv\>}_{=\,0}> 0,\qquad\text{for all $v\in\Bbb R^n\setminus\{0\}$}.$$
Note that $\<v,Tv\>=0$ because $\<v,Tv\>=\<T^\top v,v\>=\<-Tv,v\>=-\<v,Tv\>$.
Example. Take the positive definite identity matrix $I$ and the skew-symmetric matrix
$$S=\begin{pmatrix} \phantom+0 & 1 \\ -1 & 0 \end{pmatrix}.$$
From this we obtain the positive definite but not symmetric matrix
$$A=I+S=\begin{pmatrix} \phantom+1 & 1 \\ -1 & 1 \end{pmatrix}.$$
However, the eigenvalues are no longer real (they are $1\pm i$), hence no longer positive in the usual sense. They are however still located on the positive half-plane. Anyway, non-real eigenvalues are not a problem for the defining property, as we have $\<v,Av\>=v_1^2+v_2^2> 0$.
Positive definite matrices by eigenvalues
If you define positive definite matrix by having positive eigenvalues only, then there are also non-symmetric examples. E.g. take
$$A=\begin{pmatrix} \phantom+1 & 0 \\ -1 & 2 \end{pmatrix}$$
which is not symmetric, but has positive eigenvalues $1$ and $2$ only.
You can find more such matrices in the following ways:
Method 1. Choose a basis $v_1,...,v_n$ of $\Bbb R^n$, but no orthogonal basis. In the above example I chose $(1,0)$ and $(1,1)$. Then choose $n$ different eigenvalues $\lambda_1,...,\lambda_n> 0$. Let $V=(v_1,...,v_n)^\top$ be the matrix with the $v_i$ as its rows, and $D=\mathrm{diag}(\lambda_1,...,\lambda_n)$. Then the matrix $$A=VDV^{-1}$$ is non-symmetric and has only positive eigenvalues $\lambda_1,...,\lambda_n$. That it is not be symmetric can be seen as follows: the $v_i$ will be the eigenvectors of your matrix $A$. But symmetric matrices will have an orthogonal basis of eigenvectors, while our matrix will not (because we have chosen them this way). So it cannot be symmetric.
Method 2. Choose a non-diagonal upper/lower triangluar matrix (only non-zero values above/below the diagonal). If you put only positive values on the diagonal, then this matrix will have only positive eigenvalues (as these are exactly the values on the diagonal), but it is obviously not symmetric. By putting sufficiently large values off-diagonal, we can have positive eigenvalues, while not satisfying $\<v,Av\>>0$ for all $v\in\Bbb R^n\setminus\{0\}$. Choose e.g. $$A=\begin{pmatrix} \phantom{+10}1 & 0 \\ -100 & 2 \end{pmatrix}.$$ This matrix has again eigenvalues $1$ and $2$, but $\<v,Av\>=v_1^2+2v_2^2-100v_1v_2$, hence $v=(1,1)$ gives $\<v,Av\>=-97<0$.
Recall that any matrix can be expressed as the sum of a symmetric part and a skew part as follow
$$A=\overbrace{\frac12(A+A^T)}^{symmetric}+\overbrace{\frac12(A-A^T)}^{skew}$$
and since the quadratic form for the skew part is equal to $0$ we have that the positive definiteness depends solely upon the symmetric part.
Sometimes symmetry is required in the definition. If one removes the symmetry requirement, then $$ \begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix} $$ is positive definite, but not symmetric.