Consider the polynomial $p_{0}(x,y) = yf(x,y) - x \not\equiv 0$. The problem statement implies that $p_{0}(x,y) = 0$ for all pairs $(x,y)$ with $x, y\leq n, x+y\leq n+1$. Plugging in $y=1$, this gives that the polynomial $p_{0}(x,1)-x$ in $x$ has roots $1,2,\dots, n$. However, $\deg (p_{0}(x,1) -x) \leq \deg_{x} f(x,y) < n$, hence $p_{0}(x,1)-x$ must be the zero polynomial. In particular this implies that $y-1 \mid yp_{0}(x,y)-x$, so $y-1\mid p_{0}(x,y)-x$ (therefore $\deg_{x} f(x,y) < n-1$ and we can rewrite $p_{0}(x,y) - x = (y-1)p_{1}(x,y)$ for some polynomial $p_{1}$. Now plugging in $y=2$ implies that $p_{1}(x, 2) = 0$ for all $x=1, 2, \dots, n-1$, but $\deg p_{1}(x,2) < n-1$, so $p_{1}(x,2) \equiv 0$. This shows that $p_{1}(x,y) = (y-2)p_{2}(x,y)$, which once again reduces the highest possible degree of $x$ in $f(x,y)$. Continuing onward, we get to: \[yp_{0}(x,y) - x = p_{n-1}(x,y)\prod\limits_{i=1}^{n-1} (y - i)\]Due to the $-x$ on the left hand side, $p_{n-1}(x,y)$ can't be constant, hence due to the degree condition, $p_{n-1}(x,y)$ must be linear in $x$. Comparing the constant coefficients in front of $x$ on both sides implies that \[p_{n-1}(x,y) = \frac{(-1)^{n-1}}{(n-1)!}x+p_{n}(y)\]for some linear polynomial $p_{n}y$. The only bit of information we haven't used yet is the fact that $f(1, n) = \frac{1}{n}$, which gives $p_{n}(n) = \frac{(-1)^{n}}{(n-1)!}$. However, $p_{n}(y)$ doesn't have a constant term as $yp_{0}(x,y)-x$ doesn't have a constant term, so $p_{n}(y) = \frac{(-1)^{n}}{n!}$. Finally, we determined that \[yp_{0}(x,y) - x = \frac{(nx - y)}{n!}\prod\limits_{i=1}^{n-1} (i-y).\]To find $f(0,0)$, we need to compute the constant coefficient in front of $y$ on the right-hand side, which is equal to $\frac{1}{n!}\prod\limits_{i=1}^{n-1} i = \frac{1}{n}$, so $f(0,0) = \frac{1}{n}$.