* The function $g(x) = f(x, 1)$ is bijective.
Assume that $a, b$ are two positive numbers with $f(a, 1) = f(b, 1)$.
By comparing $P(a, 1, 1)$, $P(b, 1, 1)$ we obtain
$$a^2 f(a, 1) = f(f(a, 1), 1) = f(f(b, 1), 1) = b^2 f(b, 1) \implies a = b.$$
So $f(a, 1)$ is injective. Also
$$P(1, y, 1) : f(f(1, y), 1) = y^2 f(1, 1).$$
The RHS of the above equation can be any positive real number, so $f(a, 1)$ is surjective.
* The function $h(x) = f(1, x)$ is bijective.
Note that for any positive real number $t$, we have
$$P\left(1, \underbrace{\sqrt{\frac{f(t, 1)}{f(1, 1)}}}_{y}, 1\right) : f(f(1, y), 1) = f(t, 1).$$
According to the previous claim, we must have $h(y) = f(1, y) = t$ so $h(x)$ is surjective. Now
$$P(1, 1, 1) : f(f(1, 1), 1) = f(1, 1) \implies f(1, 1) = 1.$$
Now if for some positive numbers $a, b$ we have $f(1, a) = f(1, b)$, by comparing $P(1, a, 1)$, $P(1, b, 1)$ we obtain
$$a^2 = f(f(1, a), 1) = f(f(1, b), 1) = b^2 \implies a = b.$$
So $h(x)$ is injective.
We have
$$P(1, y, z) : f(f(1, y), z) = y^2 f(1, z),$$
And also had
$$P(1, y, 1) : f(f(1, y), 1) = y^2.$$
So we obtain
$$f(h(y), z) = f(f(1, y), z) = y^2 f(1, z) = g(h(y)) h(z).$$
$$\forall a, z \in \mathbb{R}^{+} : f(a, z) = g(a)h(z).$$
Now using the above equation, we rewrite the first assertion $P(x, y, z)$ and get
$$g(g(x)h(y)) = x^2y^2g(x), \quad g(1) = h(1) = 1$$
Now set $y = 1$ to get $g(g(x)) = x^2g(x)$, also we had $g(h(y)) = y^2$, so we can rewrite the above equation as
$$g(g(x)h(y)) = g(g(x))g(h(y)) \quad \underset{\text{are surjective}}{\overset{g,h}{\implies}} \quad \forall x, y \in \mathbb{R}^{+} : g(xy) = g(x)g(y).$$
We can also get
$$\begin{array}{rcl} g(h(y)) = y^2 & \Longrightarrow & g(h(xy)) = x^2 y^2 = g(h(x))g(h(y)) = g(h(x)h(y)), \\ & \underset{\text{is injective}}{\stackrel{g}{\Longrightarrow}} & h(xy) = h(x)h(y). \\[1.5ex] g(g(x)) = x^2 g(x) & \Longrightarrow & g(g(h(x))) = h(x)^2 g(h(x))), \\ & \Longrightarrow & g(x^2) = x^2 h(x^2) = x^2 h(x^2), \\ & \Longrightarrow & \forall x \in \mathbb{R}^+ : g(x) = xh(x), \\ & \Longrightarrow & h(y)h(h(y)) = y^2. \end{array}$$
Now rewrite $Q(x, y)$
$$h(x + x^2h(xy)) \geq x + xyh(x)^2.$$
Set $y \to \frac{y}{x}$ to get
$$h(x + x^2h(y)) \geq x + yh(x)^2 \implies h(1 + xh(y)) \geq \frac{x}{h(x)} + yh(x).$$
Note that $h(1) = h(x)h(\frac{1}{x})$, so $h(\frac{1}{x}) = \frac{1}{h(x)}$. Set $x = \frac{1}{h(y)}$ above to get
$$h(2) \geq \frac{h(h(y))}{h(y)} + \frac{y}{h(h(y))} \geq 2\sqrt{\frac{y}{h(y)}} \implies \exists c \in \mathbb{R}^{+} : h(y) \geq cy.$$
Also since $h(xy) = h(x)h(y)$, we obtain $h(x^n) = h(x)^n$, for all positive integers $n$. Therefore
$$\begin{align*} h(y)^n &= h(y^n) \ge cy^n \implies h(y) \ge \sqrt[n]{cy} \stackrel{n \to \infty}{\Longrightarrow} h(y) \ge y \\ \implies y^2 &= h(y)h(h(y)) \ge y^2 \implies h(y) = y \implies g(x) = x^2 \\ \implies f(x,y) &= g(x)h(y) = x^2y \end{align*}$$
So $f(x, y) = x^2y$ is the only answer of the problem which is indeed a
solution. ■