Let $P(a,b,c)$ denote the first equation. Note that $P(a,a,a): f(a,a)=gcd(a,f(a,a))$ so $a\ge f(a,a)$. Combined with the second equation, $f(a,a)=a$. $P(a,ab,ab): f(a,ab)=a$ $P(a,ck,c): f(gcd(a,ck),c)=gcd(a,c)$ Define $gcd(a,k,c)=d, gcd(a,c)=dx, gcd(a,k)=dy$ (obviously $gcd(x,y)=1$ else we can stuff more stuff into $d$) Now let $a=dxya_1$, $c=dxc_1$, $k=dyk_1$ where $gcd(a_1,x)=gcd(a_1,y)=1=gcd(y,c_1)$ (The first two relations follow else we can stuff more stuff into $x,y$, the last relation is because if not, we can stuff more stuff into $d$) Then the above relation ($f(gcd(a,ck),c)=gcd(a,c)$) gives $f(dxy,dxc_1)=dx$ and thus $f(a,b)=gcd(a,b)$, which obviously works.

All possible solution is $f(a,b)=\gcd(a,b)$ which clearly work. To verify that it is the only solution, let $P(a,b,c)$ denote the assertion $f(\gcd(a,b),c)=\gcd(a,f(c,b))$. $P(a,a,a)\implies f(a,a)=\gcd(a,f(a,a))\implies f(a,a)\leq a\stackrel{f(a)\geq a}{\implies} f(a,a)=a$ $P(b,ab,ab)\implies f(b,ab)=\gcd(b,ab)=b$ $P(a,ab,b)\implies f(a,b)=\gcd(a,f(b,ab))=\gcd(a,b)$