Yes. Let $p=char(K)$.

- $K^{1/p}[x^{1/p}]\cap K((x))=K[x]$

- If a rational function $u/v\in Frac(K^{1/p}[x^{1/p}])\cap K((x))$ then $v^p\in K[x]$ so that $u v^{p-1}\in K^{1/p}[x]\cap K((x))= K[x]$

  and hence $$K((x)) \cap K(x)^{1/p}= K((x)) \cap Frac(K^{1/p}[x^{1/p}])= K(x)$$


> Let $\alpha \in K((x))\cap \overline{K(x)}$ 

Its $K(x)$-minimal polynomial is 
$Q(y)\in K(x)[y]$. If it is inseparable then $Q(y) = R(y)^p$ with $$R(y)\in K(x)^{1/p}[y]\cap K((x))[y] = K(x)[y]$$

Which contradicts the minimality of $Q$.