Functional equations What are the general solutions of the functional equations?
$$
f(x,y)+f(y,z)=\frac{1}{f(x,z)}
$$
$$
f(x,y)f(y,z)f(x,z)=1
$$
 A: You don't need to assume any differentiability (or even continuity).  In fact, the domain of $f$ need have no structure other than being a set of the form $D\times D$ for some other set $D$.  
I'll assume that the range of $f$ in each case is $\mathbb{R}$, even though the OP didn't specify that.  I suppose that, more generally, for the first equation, $f$ could take values in any field $\mathbb{F}$ (or even a division ring), and the answer might differ a little bit in that case, depending on the nature of $\mathbb{F}$.  For the second equation, perhaps one could allow $f$ to take values in a group $\mathbb{G}$ to get a more general problem.
For the first equation, setting $x=y=z$, one sees that $f(x,x)=\epsilon(x)/\sqrt{2}$ for some function $\epsilon$ that satisfies $\epsilon(x)=\pm 1$.  However, now, setting $z=x$, one has $f(x,y)+f(y,x)=\sqrt{2}\epsilon(x)=\sqrt{2}\epsilon(y)$ for all $x$ and $y$, so $\epsilon$ is constant.  Replacing $f$ by $-f$ if necessary, one can assume $\epsilon(x)\equiv1$, so that $f(x,y)+f(y,x)\equiv\sqrt{2}$.  Thus, set $f(x,y) = 1/\sqrt{2}+a(x,y)$ where $a(x,y)=-a(y,x)$.  Substituting this back into the equation with $x=y$ yields $\sqrt{2}+a(x,z)=1/\bigl(1/\sqrt{2}+a(x,z)\bigr)$, which gives $a(x,z)\bigl(a(x,z)+3/\sqrt{2}\bigr)=0$.  Thus, either $a(x,z)=0$ or $a(x,z)=-3/\sqrt{2}$.  However, the latter is not possible since then $a(z,x)=3/\sqrt{2}$, which is not allowed.  Thus, $a(x,z)=0$ for all $x$ and $z$.  Thus, $f(x,y)\equiv1/\sqrt{2}$ and $f(x,y)\equiv -1/\sqrt{2}$ are the only solutions.
Similarly, for the second equation, one has $f(x,x)^3=1$, so $f(x,x)=1$.  Setting $z=x$, yields $f(x,y)f(y,x)=1$, so $f(x,y)=1/f(y,x)$ for all $x$ and $y$.  Suppose that there is a pair $(x,y)$ such that $f(x,y) < 0$.  Then for any $z$, one has $f(x,z)f(y,z) < 0$, so either $f(x,z) < 0$ or $f(y,z) < 0$ (and not both).  Let $Y$ be the set of $z$ such that $f(x,z) < 0$ and let $X$ be the set of $z$ such that $f(y,z) < 0$.  Then $X$ and $Y$ are disjoint and nonempty and their union is everything.   If $z$ and $w$ belong to $Y$, then the equation $f(x,z)f(z,w)f(x,w)=1$ implies $f(z,w)>0$.  Similarly, if $z$ and $w$ belong to $X$, then $f(z,w)>0$.  
Letting $\epsilon(z,w)=1$ when $f(z,w)>0$ and $\epsilon(z,w)=-1$ when $f(z,w)<0$, the function $g$ defined by $g(z,w)=f(z,w)\epsilon(z,w)=|f(z,w)|$ will satisfy the same functional equation as $f$ but will be positive everywhere.  Writing $h(z,w)=\log\bigl(g(z,w)\bigr)$, one sees that $h$ satisfies the functional equation
$$
h(u,v) + h(v,w) + h(u,w)=0
$$
But this says that $h(v,w)=-h(u,v)-h(u,w)=h(w,v)$, while one already knows that $h(v,w)=-h(w,v)$.  Thus $h(v,w)\equiv0$, so $g(v,w)\equiv1$.
Thus, the solutions of the second functional equation are obtained as follows: Write the domain $D$ as a disjoint union of two sets $X$ and $Y$ and set $f(z,w) = +1$ when $z$ and $w$ both belong to either $X$ or $Y$ and $f(z,w)=-1$ otherwise. 
A: If you assume smoothness of f, both identities lead to f being constant. For the first one, take the derivative with respect to x and y. You find $$f_{xy}=0.$$ Now take the derivative with respect to x and z. This yields zero on the left, and on the right, since you already know $$f_{xz}=0,$$ you find $$f_z=0.$$ So we have $$f(x,y)=a(x)+b(y)$$ with b'=0, and f is constant.
For the second problem, let $$g=\ln f,$$ and you find $$g(x,y)+g(y,z)+g(x,z)=0.$$ Again you find $$g_{xy}=0,$$ so $$g(x,y)=a(x)+b(y),$$ and $$2a(x)+a(y)+b(y)+2b(z)=0.$$ Taking derivatives with respect to x and z, we find that a and b must be constant.
