What are the steps to solve this simple algebraic equation?

First, clear the denominator by multiplying both sides by $x$: \begin{align*} z &= \frac{100(x-y)}{x}\\ zx &= 100(x-y) \end{align*} Then move all the terms that have an $x$ in it to one side of the equation, all other terms to the other side, and factor out the $x$: \begin{align*} zx &= 100x - 100y\\ zx - 100x &= -100y\\ x(z-100) &= -100y \end{align*} Now divide through by $z-100$ to solve for $x$; you have to worry about dividing by $0$, but in order for $z-100$ to be $0$, you need $z=100$; the only way for $z$ to be equal to $100$ is if $\frac{x-y}{x}=1$, that is, if $x-y=x$, that is, if $y=0$. Since, presumably, you don't get the things for free, you can assume that $y\neq 0$ so this division is valid. You get: $$x = \frac{-100y}{z-100}.$$ Now, to get it into nicer form, use the minus sign in the numerator to change the denominator from $z-100$ to $100-z$. Then divide both the numerator and the denominator by $100$ to get it into the form you have: \begin{align*} x & = \frac{-100y}{z-100}\\ x &= \frac{100y}{100-z}\\ x &= \frac{\frac{1}{100}\left(100 y\right)}{\frac{1}{100}(100-z)}\\ x &= \frac{y}{1 - \frac{z}{100}}. \end{align*}

Added: Alternatively, following Myself's very good point, you can go "unsimplify" $\frac{x-y}{x}$ to $1 - \frac{y}{x}$, to go from $$\frac{z}{100} = \frac{x-y}{x} = 1 - \frac{y}{x}$$ to $$\frac{y}{x} = 1 - \frac{z}{100}.$$ Taking reciprocals and multiplying through by $y$ gives \begin{align*} \frac{x}{y} = \frac{1}{1 - \frac{z}{100}}\\ x = \frac{y}{1-\frac{z}{100}} \end{align*} which is probably how the particular expression you had (as opposed to $\frac{100y}{100-z}$) arose in the first place.