Answer: $2$. Equality Case: $(x,y,z)=(0,1,1)$ or $(x,y,z)=(0,1,-1)$ Proof: $\left|\dfrac yz-xz\right|\leq 1\Rightarrow \dfrac {y^2}{z^2}+x^2z^2-2xy\leq 1$. $\left|\dfrac xz+yz\right|\leq 1\Rightarrow \dfrac {x^2}{z^2}+y^2z^2+2xy\leq 1$. Then $\dfrac {y^2}{z^2}+x^2z^2+\dfrac {x^2}{z^2}+y^2z^2\leq 2\Rightarrow (x^2+y^2)(z^2+\dfrac 1{z^2})\leq 2$. By AM-GM $z^2+\dfrac 1{z^2}\ge 2$. Hence, $x^2+y^2\leq 1$. Let $f(x,y)=x^3+2y$. We are searching for the maximum value of this function while $x^2+y^2\leq 1$. Obviously $f(x,y)\leq f(|x|,|y|)$. So, we can assume $x,y\ge 0$. $f(x,y)$ is increasing while $x$ is increasing. So, we can assume $x^2+y^2=1$. $x^3+2y\leq 2\Leftrightarrow x^3+2(y^3+x^2y)\leq 2\Leftrightarrow (x^3+2y^3+2x^2y)^2\leq 4(x^2+y^2)^3\Leftrightarrow$ $x^6+4y^6+4x^4y^2+4x^3y^3+4x^5y+8y^4x^2\leq 4x^6+4y^6+12x^4y^2+12y^4x^2\Leftrightarrow$ $4x^3y^3+4x^5y\leq 3x^6+8x^4y^2+4y^4x^2\Leftrightarrow 4xy^3+4x^3y\leq 3x^4+8x^2y^2+4y^4$. By AM-GM $x^4+x^4+x^4+y^4\ge 4x^3y$. Then it suffices to prove that $8x^2y^2+3y^4\ge 4xy^3$. $8x^2y^2+3y^4\ge 4xy^3\Leftrightarrow 8x^2+3y^2\ge 4xy\Leftrightarrow 2(x-y)^2+6x^2+y^2\ge 0$, which is correct.

BarisKoyuncu wrote: Answer: $2$. Equality Case: $(x,y,z)=(0,1,1)$ or $(x,y,z)=(0,1,-1)$ Proof: $\left|\dfrac yz-xz\right|\leq 1\Rightarrow \dfrac {y^2}{z^2}+x^2z^2-2xy\leq 1$. $\left|\dfrac xz+yz\right|\leq 1\Rightarrow \dfrac {x^2}{z^2}+y^2z^2+2xy\leq 1$. Then $\dfrac {y^2}{z^2}+x^2z^2+\dfrac {x^2}{z^2}+y^2z^2\leq 2\Rightarrow (x^2+y^2)(z^2+\dfrac 1{z^2})\leq 2$. By AM-GM $z^2+\dfrac 1{z^2}\ge 2$. Hence, $x^2+y^2\leq 1$. Let $f(x,y)=x^3+2y$. We are searching for the maximum value of this function while $x^2+y^2\leq 1$. Obviously $f(x,y)\leq f(|x|,|y|)$. So, we can assume $x,y\ge 0$. hence another continuation below $$x^3+2y\le x^2+2\sqrt{1-x^2}\le x^2+1+1-x^2=2$$done.

$$\left|\dfrac yz-xz\right|\leq 1\Rightarrow \dfrac {y^2}{z^2}+x^2z^2-2xy\leq 1,\left|\dfrac xz+yz\right|\leq 1\Rightarrow \dfrac {x^2}{z^2}+y^2z^2+2xy\leq 1$$Then $$\dfrac {y^2}{z^2}+x^2z^2+\dfrac {x^2}{z^2}+y^2z^2\leq 2\Rightarrow (x^2+y^2)(z^2+\dfrac 1{z^2})\leq 2$$By AM-GM $z^2+\dfrac 1{z^2}\ge 2$. Hence, $x^2+y^2\leq 1$ $$x^3+2y\le x^2+2\sqrt{1-x^2}\le x^2+1+1-x^2=2$$Equality Case:$(x,y,z)=(0,1,1)$ or $(x,y,z)=(0,1,-1)$

$x^2+y^2\le 1 \to x,y\in [-1,1]$ $x^3+2y\le x^2+y^2+1\le 2$

BarisKoyuncu wrote: Let $x,y,z$ be real numbers such that $$\left|\dfrac yz-xz\right|\leq 1\text{ and }\left|yz+\dfrac xz\right|\leq 1$$Find the maximum value of the expression $$x^3+2y$$ Let $x,y>0$; $|x-2y| \leqslant \frac{1}{\sqrt{x}}$ and $|y-2x| \leqslant \frac{1}{\sqrt{y}}$. Find maximum: $$P=x^2+2y$$