Yes. Let $p=char(K)$ and $\alpha\in \overline{K(x)}\cap K((x))$ assumed to be inseparable over $K(x)$.
Let $L= K^{1/p^\infty}$ which is perfect. If $\alpha$ is inseparable over $L(x)$ then $\alpha$'s monic $L(x)$-minimal polynomial is $f(y)=g(y^p)=h(y)^p$ with $h(y)\in L(x)^{1/p}[y]=L(x^{1/p})[y]$ so $$[L(x^{1/p},\alpha):L(x)] = p \deg(h) =[L(x,\alpha):L(x)]$$ ie. $x^{1/p}\in L(x,\alpha)$.
This gives that $x^{1/p}\in L((x))[\alpha]= L((x))$ which is a contradiction.
Whence $\alpha$ is separable over $L(x)$ $\implies f(y)$ is separable.
There is a finite extension $E/K$ such that $f(y)\in E[y]$. Take a basis $E=\bigoplus_{j=1}^q b_j K$ with $b_1=1$. We get that $$E(x)=\bigoplus_{j=1}^q b_j K(x),\qquad E((x))=\bigoplus_{j=1}^q b_j K((x))$$ $$f(y)=\sum_{j=1}^q b_j f_j(y), \qquad f_j(y)\in K(x)[y]$$
$f(\alpha)=\sum_{j=1}^q b_j f_j(\alpha) = 0$ in $E((x))$ gives that $$f_1(\alpha)=0$$ $f_1(y)\in K(x)[y]$ being monic and of degree $= \deg f$ it must be that $f(y)=f_1(y)$.
$f_1$ is in $K(x)[y]$, separable, irreducible, which proves that $\alpha$ is in fact separable over $K(x)$.