For all real numbers $x$ and $y$ such that $xy\neq0$, $f(x,y)=x^2-\frac{1}{y}$. If $k>1$, which of the following must be true?
I. $f(k,k)\cdot f(k,-k)>0$
II. $f\left(f\left(2,\frac{1}{k}\right),\frac{1}{k}\right)>f\left(2,\frac{1}{k}\right)$
III. $f\left(k,\frac{1}{k}\right)+f\left(\frac{1}{k},k\right)>0$

This one is tough. I got I and III but couldn't figure out II. The nested function is confusing.

For II, note that f(2,1/k) = 4 - k. Since k>1, this is less than 3. Then f(f(2,1/k),1/k) = (4-k)^2 - k. Need to compare with 4-k. Let's test k=2: f(2,1/2)=3, then f(3,1/2)=9-2=7 >3. For k=3: f(2,1/3)=4-3=1, then f(1,1/3)=1-3=-2, which is not >1. So II fails for k=3. Thus II is not always true.

I is straightforward: f(k,k) = k^2 - 1/k >0 since k>1. f(k,-k) = k^2 + 1/k >0. Product positive. III: f(k,1/k) = k^2 - k, f(1/k,k) = 1/k^2 - 1/k. Sum = k^2 + 1/k^2 - (k+1/k). For k>1, is this always >0? Check k=2: 4+0.25-2.5=1.75>0. k=1.1: 1.21+0.826-1.1-0.909=0.027>0. Seems true. So I and III only, answer D.