Theory
This is a variation of Rice's theorem, which says that if $C$ is a non-empty and proper subclass of the class of all recursive functions, then the set $\{x : \phi_x\in C\}$ is not decidable. Hence, the function
$$g(x)=
\begin{cases}
1,\text{ if }\phi_x\in C\\
0, \text{ else}
\end{cases}$$
is not computable.
A short application
We put $C$ to be the subclass of all functions $\phi_x$ such, that $h(x)\downarrow\Rightarrow \exists n.\phi_x(n)\downarrow$. Take
$$k(x)=
\begin{cases}
0,\text{ if } x>0\\
\uparrow,\text{ else }
\end{cases}$$
We note that $k\in C$, because $k(1)\downarrow$, so $C$ is non-empty. Also, it is a proper subclass, since there is a partial recursive function that does not belong in $C$, namely the nowhere defined function $\Omega(x)=\uparrow$.
You can define
$$
g(x)=
\begin{cases}
1,\text{ if }h(x)\downarrow\\
0,\text{ else }
\end{cases}
=
\begin{cases}
1,\text{ if }\phi_x\in C\\
0,\text{ else }
\end{cases}
$$
Then, by Rice's theorem, $g(x)$ is not computable. Then $h(x)$ is not computable either, because if it was, then $g$ would be computable too. But it's not because Rice's theorem says so.
A variant
If you don't want to use Rice's theorem, you can use a variation of its proof.
Here we put $C$ to be the subclass of all partial recursive functions for which $\exists n.\phi_x(n)\downarrow$. So actually $C$ is the same as above and we already have found a $k\in C$.
So now, we note that
$$h(x)=
\begin{cases}
\mu y.\phi_x(y)\downarrow,\text{ if }\exists n.\phi_x(n)\downarrow\\
\uparrow, \text{ else}
\end{cases}
=
\begin{cases}
\mu y.\phi_x(y)\downarrow,\text{ if }\phi_x\in C\\
\uparrow, \text{ else}
\end{cases}$$
is quite similar to $g$ defined above. We assume it is computable. The proof of Rice's theorem uses a computable function $f(x,y)$ defined with the help of some computable $k\in C$. Then, he does some manipulations (which involve the S-n-m theorem) and concludes that $g$ composed with another decidable function decides the halting problem. This is a contradiction since the halting problem is not decidable, hence $g$ is not computable. We will use a similar method for $h$ instead of $g$.
So, let
$$f(x,y)=
\begin{cases}
k(y), \text{ if }\phi_x(x)\downarrow\\
\uparrow,\text{ else}
\end{cases}
=
\begin{cases}
0, \text{ if }y>0\text{ or }\phi_x(x)\downarrow\\
\uparrow,\text{ else}
\end{cases}$$
Note that this is the function on your post and it is computable. The role of $y>0$ is that we need to use some function from subclass $C$. And we found such a function $k\in C$, which uses the case $y>0$. By the S-n-m theorem there is a primitive recursive (meaning, computable) function $l(x)$ such, that $f(x,y)=\phi_{l(x)}(y)$. Let's see what's happening when trying to compute $h(l(x))$ which is assumed to be computable.
$$
h(l(x))=
\begin{cases}
\mu y.\phi_{l(x)}(y)\downarrow,\text{ if }\exists n.\phi_{l(x)}(n)\downarrow\\
\uparrow, \text{ else}
\end{cases}
=
\begin{cases}
\mu y.f(x,y)\downarrow,\text{ if }\exists n.f(x,n)\downarrow\\
\uparrow, \text{ else}
\end{cases}
=\\=
\begin{cases}
\mu y.f(x,y)\downarrow,\text{ if }\exists n.(f(x,n)=k(n))\\
\uparrow, \text{ else}
\end{cases}
=
\begin{cases}
\mu y.f(x,y)\downarrow,\text{ if }\phi_x(x)\downarrow\\
\uparrow, \text{ else}
\end{cases}
=\text{HP}
$$
So you have in your hands a function that decides the halting problem, namely $h\circ l$. This is a contradiction. So, $h\circ l$ is not computable. But you know that $l$ is, so the wrong assumption leading you to contradiction was that $h$ is computable.
