smc

Plug in $t=2$ and test each expression (if the relation works, then both sides are equal). After testing, options A, B, and D are out, but for option C, both sides are equal. To check that C is a valid option, substitute the values of $x$ and $y$ and use exponent properties. $$x^y=(t^{\frac{1}{t-1}}{)}^{t^{\frac{t}{t-1}}}=t^{\frac{1}{t-1}\cdot t^{\frac{t}{t-1}}}$$ $$y^x=(t^{\frac{t}{t-1}}{)}^{t^{\frac{1}{t-1}}}=t^{\frac{t}{t-1}\cdot t^{\frac{1}{t-1}}}$$ In the second equation, the exponent can be rewritten as $\frac{1}{t-1}\cdot t\cdot t^{\frac{1}{t-1}}$. Since $\frac{t-1}{t-1}=1$, the exponent can be simplified as $\frac{1}{t-1}\cdot t^{\frac{t-1}{t-1}}\cdot t^{\frac{1}{t-1}}=\frac{1}{t-1}\cdot t^{\frac{t}{t-1}}$. Since the base and exponent of both $x^y$ and $y^x$ are the same, we can confirm that $\boxed{y^x=x^y}$.