Isomorphism between projective varieties $\mathbf{P}^{1}$ and a conic in $\mathbf{P}^{2}$

1) The isomorphism $\phi$ is between $\mathbb P^1$ and a conic $Y\subset\mathbb P^2$, not a cone.
(This conic is related to the cone in $\mathbb A^3$ with the same equation, but you should not confuse these two algebraic varieties).
2) As in all categories there are morphisms between varieties but you must keep in mind a most important point: those morphisms are defined locally and there is no hope that a morphism can be given by a single "formula": see the "Morality" below.
3) Your question beautifully illustrates the difficulty in 2) and allows for a pedagogical illustration of how to correctly define the inverse morphism $\psi=\phi^{-1}:Y\to \mathbb P^1$, namely:

a) The conic $Y$ has an open (in the Zariski topology!) covering $\{U,V\}$, where $U=Y\setminus \{[-1:0:1]\}$ consists of the points $p=[x:y:z]\in Y$ satisfying $y\neq 0$ or $x+z\neq0$ and $V=Y\setminus \{[1:0:1]\}$ consists of the points $q=[x:y:z]\in Y$ satisfying $z-x\neq 0$ or $y\neq 0$.
b) We have a morphism $\psi_ U:U\to \mathbb P^1:[x:y:z]\mapsto [y:x+z]$
and a morphism $\psi_ V:V\to \mathbb P^1:[x:y:z]\mapsto [z-x:y]$
c) The morphisms $\psi_ U$ and $\psi_ V$ coincide in the (huge!) open intersection $U\cap V\subset Y$and thus together define the required isomorphism $\psi=\phi^{-1}:Y\to \mathbb P^1$.
d) Notice that our maps $\psi_ U$ and $\psi_ V$ are morphisms (and thus continuous) but to state merely that they are continuous in the Zariski topology is worthless: for example any set-theoretic bijection between curves is a homeomorphism but is absurdly far from being an isomorphism in general.

Morality
The morphism $\psi$ is a perfectly defined inverse of $\phi$, but it is impossible to write it in the form $\psi([x:y:z])=[f(x,y,z):g(x,y,z):h(x,y,z)]$ with homogeneous polynomials of the same degree $f(x,y,z),g(x,y,z), h(x,y,z)\in k[x,y,z]$.