Not $x = y = z$

From the hypothesis, we get $xy+yz+zx=\dfrac{1}{4}(x+y+z)^2$ .

So, $P=\dfrac{x^3+y^3+z^3}{2(x+y+z)(xy+yz+zx)}=\dfrac{2(x^3+y^3+z^3)}{(x+y+z)^3}$

$\qquad \;\;=2\left[\left(\dfrac{x}{x+y+z}\right)^3+\left(\dfrac{y}{x+y+z}\right)^3+\left(\dfrac{z}{x+y+z}\right)^3\right]$

Let $a=\dfrac{x}{x+y+z}; b=\dfrac{y}{x+y+z}; c=\dfrac{z}{x+y+z}$ , we have $\left\{\begin{array}{l}a+b+c=1\\ab+bc+ca=\dfrac{1}{4}\end{array}\right.$

Or $\left\{\begin{array}{l}b+c=1-a\\bc=a^2-a+\dfrac{1}{4}\end{array}\right.$

From the inequality $(b+c)^2\ge 4bc$ , we get $0\le a\le\dfrac{2}{3}$ .

We have:

$P=2(a^3+b^3+c^3)$

$\quad =2(a^3+(b+c)^3-3bc(b+c)$

$\quad =2\left(a^3+(1-a)^3-3\left(a^2-a+\dfrac{1}{4}\right)(1-a)\right)$

$\quad =6a^3-6a^2+\dfrac{3}{2}a+\dfrac{1}{2}$

$\quad =\dfrac{11}{18}+\dfrac{1}{18}(3a-2)(6a-1)^2\le\dfrac{11}{18}$

So, $\max P=\dfrac{11}{18}\boxed{\approx 0.611}$ .

The equality holds when $(x,y,z)=(4k,k,k)$ or any permutations.

Let's use the following notation: $S_1=x+y+z$ , $S_2=xy+xz+yz$ and $S_3=xyz$ . Then, from the hypothesis we know that $S_1^2=4S2$ , and the expression we want to maximize is:

$P=\dfrac{x^3+y^3+z^3}{(x+y+z)(x^2+y^2+z^2)}=\dfrac{S_1^3-3S_1S_2+3S3}{2S_1S_2}$

Then, simplify it using the hypothesis, in a way such that we only need $S_1$ and $S_3$ :

$P=\dfrac{1}{2}+\dfrac{6S_3}{S_1^3}$

Now, consider a polynomial with roots $x,y,z$ , of course it is $P(x)=x^3-S_1x^2+S_2x-S_3$ . Then, for the roots to be real, the discriminant of this polynomial must be $\geq 0$ , so:

$S_1^2S_2^2-4S_2^3-4S_1^3S_3+18S_1S_2S_3-27S_3^2 \geq 0$

Again, using the hypothesis, we get:

$\dfrac{1}{2}S_1^3-27S_3 \geq 0$

Hence, $\dfrac{S_3}{S_1^3} \leq \dfrac{1}{54}$

Finally, $P \leq \dfrac{1}{2}+6\left(\dfrac{1}{54}\right)=\boxed{\dfrac{11}{18}}$

Let $x+y+z=2S$ for some $S$ .

From what is given $(x+y+z)^{2}=4(xy+yz+zx)$ hence $xy+yz+zx=S^{2}, x^{2}+y^{2}+z^{2}=2S^{2}$ . Thus the denominator is $(x+y+z)(x^{2}+y^{2}+z^{2})=4S^{3}$ .

Also $(x+y-z)^{2}=4xy$ .

Consider the numerator.

$x^{3}+y^{3}+z^{3}=(x+y+z)(x^{2}+y^{2}+z^{2}-xy-yz-zx)+3xyz=2S^{3}+3xyz \leq 2S^{3}+\frac{4}{9}S^{3}$ .

The last inequality follows because

$16xyz=4z(4xy)=4z(x+y-z)(x+y-z) \leq (\frac{2(x+y+z)}{3})^{3}=\frac{64}{27}S^{3}$ by AM-GM inequality. (We can make sure all terms in the AM-GM are positive by letting $x \geq y \geq z$ without loss of generality).

Hence $\frac{x^{3}+y^{3}+z^{3}}{(x+y+z)(x^{2}+y^{2}+z^{2})} \leq \frac{2S^{3}+\frac{4}{9}S^{3}}{4S^{3}}=\frac{11}{18}$ .

However when $x=4y=4z$ we have equality, so answer is $\frac{11}{18}$ .