Note that

$\begin{array}{rl}\frac{{y}^{2}}{x+y}+\frac{{z}^{2}}{y+z}+\frac{{x}^{2}}{z+x}& =(\frac{{x}^{2}}{x+y}+y-x)+(\frac{{y}^{2}}{y+z}+z-y)+(\frac{{z}^{2}}{z+x}+x-z)\\ & =\frac{{x}^{2}}{x+y}+\frac{{y}^{2}}{y+z}+\frac{{z}^{2}}{z+x}=\frac{3}{2}\phantom{\rule{thinmathspace}{0ex}}.\end{array}$

Therefore,

$\frac{{x}^{2}+{y}^{2}}{x+y}+\frac{{y}^{2}+{z}^{2}}{y+z}+\frac{{z}^{2}+{x}^{2}}{z+x}=3\phantom{\rule{thinmathspace}{0ex}}.$

By the AM-GM Inequality, we see that

$\begin{array}{}\text{(*)}& \left(\frac{{x}^{2}+{y}^{2}}{x+y}\right)\left(\frac{{y}^{2}+{z}^{2}}{y+z}\right)\left(\frac{{z}^{2}+{x}^{2}}{z+x}\right)\le 1\phantom{\rule{thinmathspace}{0ex}}.\end{array}$

We claim that, for $a,b>0$,

$\begin{array}{}\text{(\#)}& \frac{{a}^{3}+{b}^{3}}{2}\le {\left(\frac{{a}^{2}+{b}^{2}}{a+b}\right)}^{3}\phantom{\rule{thinmathspace}{0ex}},\end{array}$

and the inequality becomes an equality if and only if $a=b$. To justify the claim, let $p(t)\in \mathbb{R}[t]$ be the polynomial

$p(t):=2\phantom{\rule{thinmathspace}{0ex}}{({t}^{2}+1)}^{3}-(t+1{)}^{3}({t}^{3}+1)\phantom{\rule{thinmathspace}{0ex}}.$

Since $p(1)=16-16=0$, we know that $(t-1)$ is a factor of $p(t)$. Now,

${p}^{\prime}(t)=12\phantom{\rule{thinmathspace}{0ex}}t({t}^{2}+1{)}^{2}-3\phantom{\rule{thinmathspace}{0ex}}(t+1{)}^{2}({t}^{3}+1)-3\phantom{\rule{thinmathspace}{0ex}}{t}^{2}(t+1{)}^{3}\phantom{\rule{thinmathspace}{0ex}},$

which satisfies ${p}^{\prime}(1)=48-24-24=0$ again. That is, $(t-1{)}^{2}$ is a factor of $p(t)$. We proceed further:

${p}^{\u2033}(t)=48\phantom{\rule{thinmathspace}{0ex}}{t}^{2}({t}^{2}+1)+12\phantom{\rule{thinmathspace}{0ex}}({t}^{2}+1{)}^{2}-6\phantom{\rule{thinmathspace}{0ex}}(t+1)({t}^{3}+1)-18\phantom{\rule{thinmathspace}{0ex}}{t}^{2}(t+1{)}^{2}-6\phantom{\rule{thinmathspace}{0ex}}t(t+1{)}^{3}\phantom{\rule{thinmathspace}{0ex}}.$

We have again that ${p}^{\u2033}(1)=96+48-24-72-48=0$, and so $(t-1{)}^{3}$ is a factor of $p(t)$. Now,

${p}^{\u2034}(t)=144\phantom{\rule{thinmathspace}{0ex}}t({t}^{2}+1)+96\phantom{\rule{thinmathspace}{0ex}}{t}^{3}-6\phantom{\rule{thinmathspace}{0ex}}({t}^{3}+1)-54\phantom{\rule{thinmathspace}{0ex}}{t}^{2}(t+1)-54\phantom{\rule{thinmathspace}{0ex}}t(t+1{)}^{2}-6\phantom{\rule{thinmathspace}{0ex}}(t+1{)}^{3}\phantom{\rule{thinmathspace}{0ex}},$

so ${p}^{\u2034}(1)=288+96-12-108-216-48=0$, whence $(t-1{)}^{4}$ is a factor of $p(t)$. Because $p$ is a monic polynomial of degree $6$, we must have

$p(t)=(t-1{)}^{4}\phantom{\rule{thinmathspace}{0ex}}({t}^{2}+\alpha t+\beta )$

for some $\alpha ,\beta \in \mathbb{R}$. With $p(0)=1$, we get $\beta =1$. As $p(-1)=16$, we conclude that $1-\alpha +\beta =1$, so that $\alpha =1$, as well. Consequently,

$p(t)=(t-1{)}^{4}\phantom{\rule{thinmathspace}{0ex}}({t}^{2}+t+1)\phantom{\rule{thinmathspace}{0ex}},$

which is a nonnegative polynomial (i.e., $p(\mathbb{R})\subseteq {\mathbb{R}}_{\ge 0}$), and the only real root of $p(t)$ is $t=1$.

Now, (#) is equivalent to

$2\phantom{\rule{thinmathspace}{0ex}}{({a}^{2}+{b}^{2})}^{3}-(a+b{)}^{3}({a}^{3}+{b}^{3})={b}^{6}\phantom{\rule{thinmathspace}{0ex}}p\left(\frac{a}{b}\right)=(a-b{)}^{4}\phantom{\rule{thinmathspace}{0ex}}({a}^{2}+ab+{b}^{2})\ge 0\phantom{\rule{thinmathspace}{0ex}},$

which is an equality iff $a=b$. Hence, the claim is established, but then we conclude that

$\left(\frac{{x}^{3}+{y}^{3}}{2}\right)\left(\frac{{y}^{3}+{z}^{3}}{2}\right)\left(\frac{{z}^{3}+{x}^{3}}{2}\right)\le 1\phantom{\rule{thinmathspace}{0ex}},$

using (#) in (*). Therefore,

$({x}^{3}+{y}^{3})({y}^{3}+{z}^{3})({z}^{3}+{x}^{3})\le 8\phantom{\rule{thinmathspace}{0ex}}.$

However, the problem statement demands that the inequality above is an equality. That is, $x=y=z$ must hold. Ergo, the only positive real solution to this system of equations is

$(x,y,z)=(1,1,1)\phantom{\rule{thinmathspace}{0ex}}.$