jmc

We can rewrite the third equation as $$f(x,x+y)=\frac{x+y}{y}\cdot f(x,y),$$or, making the substitution $t=x+y,$ $$f(x,t)=\frac{t}{t-x}\cdot f(x,t-x)$$whenever $x<t.$ In particular, if $r\ne 0$ is the remainder when $t$ is divided by $x,$ then repeatedly applying this relation, we have $$\begin{array}{rl}f(x,t)& =\frac{t}{t-x}\cdot f(x,t-x)\\ & =\frac{t}{t-x}\cdot \frac{t-x}{t-2x}\cdot f(x,t-2x)\\ & =\cdots \\ & =\frac{t}{t-x}\cdot \frac{t-x}{t-2x}\cdots \frac{r+x}{r}\cdot f(x,r)\\ & =\frac{t}{r}\cdot f(x,r)\end{array}$$since the product telescopes. Then we may compute $f(14,52)$ as follows, swapping the two arguments of $f$ as necessary using the second equation: $$\begin{array}{rl}f(14,52)& =\frac{52}{10}\cdot f(14,10)\\ & =\frac{52}{10}\cdot \frac{14}{4}\cdot f(10,4)\\ & =\frac{52}{10}\cdot \frac{14}{4}\cdot \frac{10}{2}\cdot f(4,2)\\ & =\frac{52}{10}\cdot \frac{14}{4}\cdot \frac{10}{2}\cdot \frac{4}{2}\cdot f(2,2)\\ & =\frac{52}{\cancel{10}}\cdot \frac{14}{\cancel{4}}\cdot \frac{\cancel{10}}{2}\cdot \frac{\cancel{4}}{2}\cdot 2\\ & =\boxed{364}.\end{array}$$