If $x=t^{1/(t-1)}$ and  $y=t^{t/(t-1)},t>0,t \ne 1$, a relation between $x$ and $y$ is: 
$\text{(A) } y^x=x^{1/y}\quad \text{(B) } y^{1/x}=x^{y}\quad \text{(C) } y^x=x^y\quad \text{(D) } x^x=y^y\quad \text{(E) none of these}$
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 $\tfrac{1}{t-1} \cdot t \cdot t^{\frac{1}{t-1}}$.  Since $\tfrac{t-1}{t-1} = 1$, the exponent can be simplified as $\tfrac{1}{t-1} \cdot t^{\frac{t-1}{t-1}} \cdot t^{\frac{1}{t-1}} = \tfrac{1}{t-1} \cdot t^{\frac{t}{t-1}}$.  Since the base and exponent of both $x^y$ and $y^x$ are the same, we can confirm that $\boxed{\textbf{(C) } y^x = x^y}$.