From looking at the graph of $Y$ versus $X$ on the given $X$-domain, the open interval $(-1,2)$,

you can see that $Y\in(-3,1]$ (note that the interval is open at $-3$ but closed at $Y=1$ where $X=0$). The CDF of $Y$ can then be calculated from the CDF of $X$, after calculating that from the integral, as follows: $ \eqalign{ F_X(x) &=P(X\le x)= \int_{-1}^{x}\tfrac19(1+t)^2\,dt =\tfrac1{27}\left[(1+t)^3\right]_{-1}^{x} =\tfrac1{27}(1+x)^3 \\\\ F_Y(y) &= P(Y\le y)=P(1-X^2\le y)=P(X^2\ge1-y)=P(|X|\ge\sqrt{1-y}) \\\\ &=\left\{\array{ 0&y\le-3\\ P\left(X\ge\sqrt{1-y}\right)\qquad&-3\lt y\le0\\ 1-P\left(|X|\le\sqrt{1-y}\right)\qquad&0\lt y\le1 \\1&y\ge1 }\right. \\\\ &= \left\{\array{ 0&y\le-3\\ 1-F_X\left(\sqrt{1-y}\right)\qquad&y\in(-3,0]\\ 1-F_X\left(\sqrt{1-y}\right)+F_X\left(-\sqrt{1-y}\right)\qquad&y\in(0,1] \\1&y\ge1 }\right. \\\\ &= \left\{\array{ 0&y\le-3\\ 1-\tfrac1{27}\left(1+\sqrt{1-y}\right)^3\qquad&y\in(-3,0]\\ 1-\tfrac1{27}\left[ \left(1+\sqrt{1-y}\right)^3- \left(1-\sqrt{1-y}\right)^3 \right]\qquad&y\in(0,1] \\1&y\ge1 }\right. } $ which for $y\in(0,1]$ can also be simplified a bit further using $(1\pm r)^3=1\pm3r+3r^2\pm r^3$ thus: $ (1+r)^3-(1-r)^3=2\,(3r+r^3)=2r\,(3+r^2)\qquad\implies $ $ F_Y(y)=1-\tfrac1{27}2\sqrt{1-y}~(3~+~1-y)=1-\frac{2\sqrt{1-y}\,(4-y)}{27} \quad\text{for}\quad y\in(0,1] $ or differentiated to get the PDF: $ \eqalign{ f_Y(y) &=\left\{\array{ 0 & \qquad y\le-3\quad\text{or}\quad y\gt1\\\\ \frac19+\frac{2-y}{18\sqrt{1-y}} & \qquad y\in(-3,0]\\\\ \frac{2-y}{9\sqrt{1-y}} & \qquad y\in(0,1] }\right. } $ Here is the PDF $f_X(x)$, in red, and CDF $F_X(x)$, in blue, of $X$:

and likewise for $Y$ (the PDF $f_Y(y)$ has a vertical asymptote at $1$):

Note that both CDFs are in fact (not only right- but also left-) continuous, so that it doesn't matter to which case we assign the transition points $-3$, $0$ and $1$; as @Dilip has pointed out, it could be considered better pedagogy to use left-closed, right-open intervals to emphasize that (CDF) distributions must be right-continuous.
The key step in my method is being able to replace the probabilities with differences of the (cumulative) distribution function for $X$ in the middle bracketed RHS above during the derivation of $F_Y$. This technique is known as the distribution function method or method of distribution functions. Alternate methods exist, for example integrating the product of $f_X$ with the derivative of the transformation function (or for multivariate transformations, the Jacobian). There is also a method using moment generating functions.
Again starting with the graph $Y=1-X^2$ above, if we approach it from an integral involving the PDF, as you do, we can still start as I did above, up to the first bracketed RHS involving probabilities:
$ \eqalign{ F_Y(y)&=P\left(X\ge\sqrt{1-y}\right) \qquad\text{for}\qquad-3\lt y\le0\\ &=\int_{\sqrt{1-y}}^2f_X(x)\,dx =\int_{\sqrt{1-y}}^2\tfrac19(1+x)^2\,dx\\ &=\tfrac1{27}\left[(1+x)^3\right]_{\sqrt{1-y}}^2 =1-\tfrac1{27}\left(1+\sqrt{1-y}\right)^3 } $ as above, and similarly for $y\in(-3,0]$. The key insight is still that $ Y=1-X^2 \le y \iff X^2 \ge 1-y \iff |X| \ge \sqrt{1-y}, $ which then must be handled seperately, depending on the sign of $x$.