I know it is true $\mathbb R(x)\otimes_{\mathbb R} \mathbb C=\mathbb C(x)$ because $\mathbb C/\mathbb R$ is algebraic. So we can always multiply by another polynomial to make the coefficients of any $f\in\mathbb C[x]$ into $\mathbb R$.
We prove the general case where $L/K$ is algebraic.
If $f$ is a monic polynomial over $L$, then we can factor into irreducible monic factors. If we know how to make each factors into $K$, then we are done. Now assume $f$ is irreducible over $L$. Then $L(a)/L$ is a finite extension where $a$ is a root of $f$. Now write down a (minimal) polynomial $h$ for $a$ for the extension $K(a)/K$. Now, $h$ also as polynomial over $L$ has a root $a$, but $f$ is the minimal polynomial. Then $f\mid h$ in $L[x]$. Now for every $f/g\in L(x)$, we can multiply by some $h$ to assume the denominator is in $K[x]$. Then every polynomial is of the form "$L[x]/K[x]$". Hence, $K(x)\otimes_KL\to L(x)$ is surjective.
Is it true that $\mathbb R(x,y)\otimes_{\mathbb R} \mathbb C=\mathbb C(x,y)$? Or for more variables.
Yes. If $L/K$ is any algebraic field extension, then the canonical homomorphism of $L$-algebras $$K(x_1,\dotsc,x_n) \otimes_K L \to L(x_1,\dotsc,x_n)$$ is an isomorphism. This follows by induction on $n$: $$\begin{align*} K(x_1,\dotsc,x_{n+1}) \otimes_K L &= K(x_1,\dotsc,x_n)(x_{n+1}) \otimes_{K(x_1,\dotsc,x_n)} K(x_1,\dotsc,x_n) \otimes_K L \\ &= K(x_1,\dotsc,x_n)(x_{n+1}) \otimes_{K(x_1,\dotsc,x_n)} L(x_1,\dotsc,x_n) \\ &= L(x_1,\dotsc,x_n)(x_{n+1}) = L(x_1,\dotsc,x_{n+1}) \\ \end{align*}$$ We have used the special case $n=1$ (that you apparently already know) as well as the fact that $L(x_1,\dotsc,x_n)$ is algebraic over $K(x_1,\dotsc,x_n)$.