$$\dfrac{1}{x^{y+1}} = y$$
Evaluate
$$\dfrac x y $$
We know that
$$\dfrac{1}{x^{y+1}} = {x^{-(y+1)}}$$
Which yields
$$x^{-y-1} = y$$
Multiplying the both sides by $-1$
$$x^{y+1} = \dfrac 1 y$$
Multiplying the both sides by $x$
$$x^{xy+1} = \dfrac x y$$
Your final step is not correct. $a\cdot a^b=a^1\cdot a^b = a^{1+b}$. Multiplying numbers with the same base is additive in the exponent, not multiplicative. So, $x\cdot x^{y+1} = x^1\cdot x^{y+1} = x^{1+y+1} = x^{y+2}$.
Putting this all together:
$$\dfrac{x}{y} = x^{y+2}$$
In terms of a single variable, you have
$x=\left(\dfrac{1}{y}\right)^{\tfrac{1}{y+1}}$
So, that gives you:
$$\dfrac{x}{y} = \left(\dfrac{1}{y}\right)^{\tfrac{y+2}{y+1}}$$