6.174. $\frac{z}{z+1}-2 \sqrt{\frac{z+1}{z}}=3$.
Solution.

Domain of definition: $\frac{z+1}{z}>0 \Leftrightarrow z \in(-\infty ;-1) \cup(0 ; \infty)$.

Let $\sqrt{\frac{z+1}{z}}=y$, where $y>0$. The equation in terms of $y$ becomes $\frac{1}{y^{2}}-2 y-3=0 \Leftrightarrow 2 y^{3}+3 y^{2}-1=0 \Leftrightarrow 2 y^{3}+2 y^{2}+y^{2}-1=0 \Leftrightarrow$ $\Leftrightarrow 2 y^{2}(y+1)+(y-1)(y+1)=0 \Leftrightarrow(y+1)\left(2 y^{2}+y-1\right)=0, y+1=0$, from which $y_{1}=-1$, or $2 y^{2}+y-1=0$, from which we find $y_{2}=-1, y_{3}=\frac{1}{2}$; $y_{1}=y_{2}=-1<0$ is not suitable.

Then $\sqrt{\frac{z+1}{z}}=\frac{1}{2} \Leftrightarrow \frac{z+1}{z}=\frac{1}{4}$, from which $z=-\frac{4}{3}$.

Answer: $z=-\frac{4}{3}$.