<s>[INST] Determine all positive integer solutions for $\frac 1x + \frac 1y + \frac 1z + \frac 1{xy} + \frac1{yz} + \frac 1{xz} + \frac 1{xyz} = 1$ I need to determine all positive integer solutions for the equation: 
$$\frac 1x + \frac 1y + \frac 1z + \frac 1{xy} + \frac1{yz} + \frac 1{xz} + \frac 1{xyz} = 1.$$
This is how I have tried to do it: 
Mulitiplied both sides by $xyz$ to get
$$yz+xz+xy+z+x+y+1=xyz.$$
Factor it:
\begin{align}
x(y+z+1-yz)+yz+y+z &= -1   \\
x(y(1-z)+z+1)+y(1+z)+z &= -1
\end{align}
If $z=0$, we get
\begin{align}
x(y+1)+y=-1 &\iff xy+x+y=-1   \\
&\iff (x+1)(y+1)=0,
\end{align}
which gives us $x=y=-1$.
Is this all positive integer solutions? Or have I missed something?
EDIT: I am stupid. 
New attempt. 
If $z=1$, I get $2x+2y=-2 \iff x+y=-1$.
Then there is no solutions of positive integers for both $x$ and $y$ at the same time.  
If I try for $z=2$, I get
\begin{align}
x(y(1-2)+2+1)+y(1+2)+2=-1 &\iff x(3-y)+3y+2=-1   \\
&\iff 3x+3y+3-xy=0
\end{align}
and I won't get a solution where all the variables are positive integers. 
 [/INST] Assume, without loss of generality, that $x\leq y\leq z$. We see from the original equation that $x>1$ (since $x = 1$ means $\frac1x + \cdots > 1$). At the same time, we must have $x<4$, as otherwise the sum is clearly less than $1$.
So, where does $x = 2$ actually take us? We insert and get
$$
\frac12 + \frac1y + \frac1z + \frac1{2y} + \frac1{2z} + \frac1{yz} + \frac1{2yz} = 1\\
\frac3{2y} + \frac3{2z} + \frac3{2yz} = \frac12\\
3z + 3y + 3 = yz\\
12 = yz - 3y - 3z + 9 = (y-3)(z-3)
$$
Since $y$ and $z$ are integers, this is an easy solve.
What about $x = 3$? We get
$$
\frac13 + \frac1y + \frac1z + \frac1{3y} + \frac1{3z} + \frac1{yz} + \frac1{3yz} = 1\\
\frac4{3y} + \frac4{3z} + \frac4{3yz} = \frac23\\
4z + 4y + 4 = 2yz\\
2z + 2y + 2 = yz\\
6 = yz - 2y-2z + 4 = (y-2)(z-2)
$$
which, again, is an easy solve using the fact that $y, z$ are integers.
</s>