Prove existence of non-computable double argument function

Sure. Let the function's range be $\{0,1\}$, such that for each $x\in\mathbb{N}$, $f(x,y)=1$ for exactly one value of $y$; call it $y(x)$, and assume it is strictly increasing with $x$. (In either words, $f$ is the graph of an increasing function.) Clearly $f(x,y)$ is computable for each fixed $x$ (it's all zeroes and a single $1$, the location of which can be specified in a finite number of bits), and $f(x,y)$ is also computable for each fixed $y$ (it's either all zeroes or again all zeroes and a single $1$). But $f(x,y)$ as a function on $\mathbb{N}^2$ is only computable if $y(x)$ is computable. And certainly we can choose $y(x)$ to be a strictly increasing non-computable function (exercise for the reader).