When you do

y = z^2/2;
fz[z_] := y;
fz[3]

the delayed evaluation f[z_]:=y means that the right-hand side is memorized verbatim. You can check with ?fz

When you then call f[3], the replacement z->3 is done in the frozen y, resulting in no replacement at all, since the verbatim y does not contain z.

After this useless replacement y/.z->3 is done, the result is unfrozen, and the outside rule y=z^2/2 gets finally applied.

Hi,

sorry, I do not get the question. Why do you say that fz[3] does not evaluate? Doesn't Mathematica do exactly what you would expect it to do?

By definition fz does not depend on $z$. So I would expect it to evaluate to $y$. But input 18 defines $y$ as $\frac{z^2}{2}$. So after fz evaluates to $y$ this is substituted by $\frac{z^2}{2}$. So Mathematica's result is as expected. I understand that this is not the result you expect. Would you have expected something like this?

ClearAll["Global`*"]
y[z_] := Evaluate[Integrate[z, z]]
fz[z_] := y[z]
fz[3]

This evaluates to $\frac{9}{2}$. By using the delayed evaluations I have changed to order of evaluations.

Cheers,

Marco