System follows: $$ y'=\frac{y^2}{z-x}; z'=y+1$$
I was found the 2 ways. The both are wrong
1) $$z = x + \frac{y^2}{y'}; z'=1+\frac{2yy'^2-y^2y''}{y'^2}=y+1; => (p(y) = y')=> yp(yp'+p)=0;$$ $$ y=0, y' = \frac{c}{y}<=>y=C e^{\frac{c}{y}}$$ it looks like an answer, but there is must be x instead y in power of e.
There was my fault $$ yp'-p=0; \frac{p'}{p}=\frac{1}{y}; p=cy=y'; \frac{dy}{y}=c dx; y=C e^{cx};$$ $$ z'=C e^{cx} + 1; z = \frac{C}{c} e^{cx} + x + const$$ There was not enough persistence of myself. Thanks for Mike
Answer is $$y=C e^{cx}; z = \frac{C}{c} e^{cx} + x + const$$
2) $$y=z'-1, y'=z''$$ $$(z-x)z''=(z'-1)^2$$ $$x=e^t, z=ue^{mt}, (m-1)=0$$ $$z'=(u+u'), z'' = e^{-t}(u'+u'')$$ $$ p(u)=u', pp'=u''$$ $$(u-1)(p+pp')=(u-1+p)^2$$ $$p'-1=\frac{u-1}{p}+\frac{p}{u-1}$$ I stuck here...
A search for my fault of the second way is open
PS: tag Филипов решение 1142
Well, since no one else has responded, I'll try taking a stab at it. Let me try reworking your first attempt from here
$$2yy'^2-y^2y''=yy'^2$$ $$yy'^2-y^2y''=0$$
I feel I'm missing something simple here, but I'll try your substitution
$$\frac{dy}{dx}=p$$ $$\frac{d^2y}{dx^2}=\frac{dp}{dx}=\frac{dp}{dy}\frac{dy}{dx}=p'p$$ $$yp^2-y^2pp'=yp(p-yp')=0$$
This seems to leave 3 possibilities. $p=0$ does not appear to work. It may have been introduced when the derivative of $z$ was taken. $y=0$ yields $z=x+k$. The last possibility is
$$yp'-p=0$$ $$\frac{p'}y-\frac p{y^2}=0$$ $$\frac py=k,p=\frac{dy}{dx}=ky$$ $$\ln y=kx+c,y=k_2e^{kx}$$ $$z'=k_2e^{kx}+1$$ $$z=\frac{k_2}ke^{kx}+x+k_3$$
Trying this back in the first equation, we have
$$\frac{y^2}{z-x}=\frac{k_2^2e^{2kx}}{\frac{k_2}ke^{kx}+k_3}$$
Again, it looks like extra solutions may have been introduced. However, with the restriction $k_3=0$, we have
$$\frac{k_2^2e^{2kx}}{\frac{k_2}ke^{kx}}=kk_2e^{kx}=y'$$