<s>[INST] If $x+\frac{1}{y+\frac{1}{z+\frac13}}=\frac{118}{51}$ find $xyz$. I'm having a difficult time figuring out how to solve this algebra problem.

If $$x+\cfrac 1 {y+\cfrac 1 {z+\cfrac 1 3}}=\frac{118}{51}$$
find $xyz$.

 [/INST] I will presume that $x,y,z$ are intended to be positive integers.
\begin{align}
\frac{118}{51} & = 2 + \frac{16}{51} = 2 + \frac 1 {\left( \frac{51}{16} \right)} = 2 + \cfrac 1 {3 + \cfrac 3 {16}} \\[15pt]
& = 2 + \cfrac 1 {3 + \cfrac 1 {\left( \frac{16} 3 \right)}} = 2 + \cfrac 1 {3 + \cfrac 1 {5 + \cfrac 1 3}}.
\end{align}
</s>