Exercise on Manifolds: Transition Maps 
Let $M\;$ be a differentiable manifold of distance $3$, $p\in M\;$ and
  two charts  $(U,φ=(x_1,x_2,x_3))\;\;,(U,ψ=(y_1,y_2,y_3))\;$ of $M$
  near $p\;$ with $φ(p)=(1,1,-2)\;$ such that:
$y_1=x_1\;,\;y_2=x_2-{x_1}^3\;,\;y_3=x_3+3x_1 {x_2}^2 \;$ in $U$
Find transition maps : $ψο{φ}^{-1}\;$ and $φο{ψ}^{-1}$

What I thought to do is: 
$ψ(x_1,x_2,x_3)=(x_1,x_2-{x_1}^3,x_3+3x_1 {x_2}^2)\;$ and $φ(y_1,y_2,y_3)=(y_1,y_2+{y_1}^3,y_3-3y_1(y_2+{y_1}^3)^2)\;$ 
So in order to compute $ψ^{-1}\;$ I write $ψ(x_1,x_2,x_3)=(x_1,x_2-{x_1}^3,x_3+3x_1 {x_2}^2)=(z_1,z_2,z_3)\;\Rightarrow \begin{cases} x_1=z_1\\x_2=z_2+{z_1}^3\\x_3=z_3-3z_1(z_2+{z_1}^3)^2 \end{cases}$
Now $φοψ^{-1}(z_1,z_2,z_3)=φ(z_1,z_2+{z_1}^3,z_3-3z_1(z_2+{z_1}^3)^2)=(z_1,z_2+2{z_1}^3,z_3-3z_1(z_2+{z_1}^3)^2-3z_1(z_2+2{z_1}^3)^2)$
In similar way I can compute $ψοφ^{-1}\;$
My question is if the above thought is correct. I feel I'm missing something... I would appreciate if somebody could help me through this. Hints or solutions other than this are of course welcome! 
Thanks in advance..

Note that $\varphi, \psi \colon U \rightarrow \mathbb{R}^3$ and you have no idea what $U$ is (it is some open subset of some manifold $M$). 
The equations 
$$y_1 = x_1, y_2 = x_2 - x_1^3, y_3 = x_3 + 3x_1 x_2^2$$ already give you (practically by definition)
$$(\psi \circ \varphi^{-1})(x_1,x_2,x_3) = (y_1,y_2,y_3) = (x_1, x_2 - x_1^3, x_3 + 3x_1 x_2^2). $$
In more details, if $q \in U$ then the coordinates of $q$ with respect to the coordinate system $\varphi$ are $\varphi(q) = (x_1(q), x_2(q),x_3(q))$. Similarly, the coordinates of $q$ with respect to the coordinate system $\psi$ are $\psi(q) = (y_1(q),y_2(q),y_3(q))$. The transition function $\psi \circ \varphi^{-1}$ eats a triple $(x_1,x_2,x_3)$ and returns the $y_i$ coordinates of the point $q = \varphi^{-1}(x_1,x_2,x_3)$. 
The meaning of the equations $y_i = f_i(x_1,x_2,x_3)$ you are given is that $y_i(q) = f_i(x_1(q),x_2(q),x_3(q))$ (the $y_i$ coordinate of $q \in U$ is related to the $x_i$ coordinates of $q$ by $f_i$). Letting $q = \varphi^{-1}(x_1,x_2,x_3)$ we get
$$ y_i(\varphi^{-1}(x_1,x_2,x_3)) = f_i(x_1(\varphi^{-1}(x_1,x_2,x_3)),x_2(\varphi^{-1}(x_1,x_2,x_3)),x_3(\varphi^{-1}(x_1,x_2,x_3))) = f_i(x_1,x_2,x_3) $$
but the left hand side is precisely the $i$-th coordinate of $\psi \circ \varphi^{-1}$.
In order to get $\varphi \circ \psi^{-1} = \left( \psi \circ \varphi^{-1} \right)^{-1}$ you need to invert $\psi \circ \varphi^{-1}$. Namely, you need to solve for the $x_i$'s in terms of the $y_i$'s. In your case,
$$ x_1 = y_1, x_2 = y_2 + x_1^3 = y_2 + y_1^3, \\
x_3 = y_3 - 3x_1x_2^2 = y_3 - 3y_1(y_2 + y_1^3)^2 $$
so
$$ (\varphi \circ \psi^{-1})(y_1,y_2,y_3) = (y_1, y_2 + y_1^3, y_3 - 3y_1(y_2 + y_1^3)^2). $$
