I have the following Boolean expression:
$$w'x'y'z + wx'y'z + xz + xyz'\tag{1}$$
Upon doing my own work, I can only get as far as:
$$zx + xy + zy'\tag{2}$$
Now, when I put the original equation into the following webpage (http://calculator.tutorvista.com/math/582/boolean-algebra-calculator.html, I've been using it to double check myself), it keeps saying the full simplification is:
$$xy + zy'\tag{3}$$
Using Truth Tables, I've compared all three equations to each other and all prove equal that I can tell, and what confuses me most is how (2) apparently simplifies to (3).
Does (2) simplify to (3)? Or am I missing some step between getting from (1) to (3)? Or is (3) simply incorrect (perhaps a glitch in the online software)?
You've already answered part of your question yourself: Since the truth tables are the same, the expressions are equivalent, so it's not a glitch in the software.
To see that $(2)$ is equivalent to $(3)$, note that for $zx$ to have an effect, both $x$ and $z$ would have to be true; but in that case $xy$ is $y$ and $zy'$ is $y'$, and $y+y'$ is true; thus $zx$ is redundant. More formally:
$$ \begin{align} xy+zy'+zx&=xy+zy'+zx(xy+zy')' \\ &=xy+zy'+zx(xy)'(zy')' \\ &=xy+zy'+zx(x'+y')(z'+y) \\ &=xy+zy'+zxy'y \\ &=xy+zy' \end{align}$$
(where I moved $zx$ to the right to make it typographically apparent that the rest stays the same).