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Solution 1. Note that $$\begin{array}{rl}\left(x+\frac{1}{y}\right)\left(y+\frac{1}{z}\right)\left(z+\frac{1}{x}\right)& =xyz+x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{xyz}\\ & =xyz+\left(x+\frac{1}{y}\right)+\left(y+\frac{1}{z}\right)+\left(z+\frac{1}{x}\right)+\frac{1}{xyz}.\end{array}$$Plugging in the given values, we have $$4\cdot 1\cdot \frac{7}{3}=xyz+4+1+\frac{7}{3}+\frac{1}{xyz}$$or $$\frac{28}{3}=xyz+\frac{22}{3}+\frac{1}{xyz}.$$Thus, $xyz+\frac{1}{xyz}=2$. Multiplying by $xyz$ and rearranging, we get $(xyz-1{)}^2=0$, so $xyz=\boxed{1}$.

Solution 2. Repeatedly substitute, in order to create an equation in a single variable. The second equation gives $y=1-\frac{1}{z}$, and the third equation gives $z=\frac{7}{3}-\frac{1}{x}$, so $$4=x+\frac{1}{y}=x+\frac{1}{1-\frac{1}{z}}=x+\frac{z}{z-1}=x+\frac{\frac{7}{3}-\frac{1}{x}}{\frac{4}{3}-\frac{1}{x}}.$$Simplifying and multiplying to clear denominators, we get the quadratic $(2x-3{)}^2=0$. Thus, $x=\frac{3}{2}$, so $z=\frac{7}{3}-\frac{1}{x}=\frac{5}{3}$ and $y=1-\frac{1}{z}=\frac{2}{5}$. Therefore, the answer is $$xyz=\frac{3}{2}\cdot \frac{2}{5}\cdot \frac{5}{3}=\boxed{1}.$$