What is the solution according to the following parameters? (equation) The equation would accept x,y and output z where
As X approaches infinity and Y approaches infinity, Z approaches 0
As X approaches infinity and Y approaches 0, Z approaches 0
As X approaches 0 and Y approaches infinity, Z approaches infinity
As X approaches 0 and Y approaches 0, Z approaches 0
I have very little knowledge in equation construction.
Thanks.
 A: There are plenty of equations satisfying the conditions you present. 
Note that an equation in the $xyz$-space is a rule 
$$f(x,y,z)=0$$
where $f$ is some conditions involving $x,y$ and $z$.
To construct one such rule $f$ we may look upon the conditions you pose and divide the $xy$-plane in disjoint pieces in which the conditions live. 
We may for example consider 
$$S_1=\{(x,y):\,|x|>1, \, |y|>1 \}$$
$$S_2=\{(x,y):\,|x|>1, \, |y|\leq 1 \}$$
$$S_3=\{(x,y):\,|x|\leq1, \, |y|> 1 \}$$
$$S_4=\{(x,y):\,|x|\leq1, \, |y|\leq 1 \}$$


*

*On $S_1$ we can define $z=f(x,y)= 1/x$, then $z$ will go to 0 as $x$ and $y$ tends to infinity. 

*On $S_2$ we can define $z=f(x,y)= y$, then $z$ will go to 0 as $x$ tends to infinity and $y$ tends to 0.

*On $S_3$ we can define $z=f(x,y)= y^8 +7$, then $z$ will go to infinity as $x$ tends to 0 and $y$ tends to infinity.

*On $S_4$ we can put $z=f(x,y)= xy$, which goes to 0 as $x$ and $y$ tends to 0.
In retrospective, we can of course take $f(x,y)=0$ everywhere except on $S_3$ where we can take $f(x,y)=y$. Also, this is just one (quiet ugly, but straight forward) way of constructing a solution. 
