We will prove that the maximal value of the given expression $P$ is $\frac{3}{16}$. Indeed, applying these inequalities
$$x^4 + y^4 \geq xy(x^2 + y^2) \text{ and } (xy + z^2)^2 \geq 4xyz^2,$$
we have
$$(x^4 + y^4)(xy + z^2)^3 \geq 4x^2y^2z^2(x^2 + y^2)(xy + z^2) \\ \geq 4x^2y^2z^2(z^2x^2 + z^2y^2 + 2x^2y^2).$$
Thus, we have
$$\frac{x^3 y^4 z^3}{(x^4 + y^4)(xy + z^2)^3} \le \frac{x^3 y^4 z^3}{4x^2 y^2 z^2 (z^2 x^2 + z^2 y^2 + 2x^2 y^2)} \\ = \frac{xy^2 z}{4(z^2 x^2 + z^2 y^2 + 2x^2 y^2)}$$
We will prove that $\sum \frac{xy^2z}{z^2x^2 + z^2y^2 + 2x^2y^2} \le \frac{3}{4}$. Put $a = xy$, $b = yz$ and $c = zx$, the left hand side becomes
$$\sum \frac{ab}{2a^2 + b^2 + c^2} \le \frac{3}{4}.$$
If $a \ge b \ge c$ then $ab \ge ac \ge bc$ and
$$\frac{1}{2c^2 + a^2 + b^2} \ge \frac{1}{2b^2 + c^2 + a^2} \ge \frac{1}{2a^2 + b^2 + c^2}.$$
Applying the rearrangement inequality, we have
$$\sum \frac{ab}{2a^2 + b^2 + c^2} \le \sum \frac{ab}{2c^2 + a^2 + b^2}.$$
By AM-GM and Cauchy-Schwarz inequalities, we obtain that
$$\begin{aligned} 4 \sum \frac{ab}{2c^2 + a^2 + b^2} & \le \sum \frac{(a+b)^2}{2c^2 + a^2 + b^2} \\ & \le \sum \left( \frac{a^2}{c^2 + a^2} + \frac{b^2}{c^2 + b^2} \right) = 3. \end{aligned}$$
$$\text{Hence, } P = \sum \frac{x^3 y^4 z^3}{(x^4 + y^4)(xy + z^2)^3} \le \frac{3}{16}.$$
The equality holds if and only if $a = b = c$ or $x = y = z$. Therefore, the maximal value of the given expression is $\frac{3}{16}$. $\square$