In response to the OP's request
As Gerry Myerson pointed out, the absolute value sign does matter, and as Michael Hardy's answer clarifies in more detail, the joint density $f_{X,Y}(x,y)$ has nonzero constant value $c$ on the region
$$\{(x,y) \colon -x < y < x, 0 < x < 1\},$$
that is, on the interior of a right triangle with vertices $(0,1), (1,1), (1,-1)$.
As I suggested in my comments on the question, sketching the $x$-$y$ plane
and marking this triangle on it is very helpful as an aid
to thought, and in this particular problem, makes the computations
very simple. In fact, it is even better
if one can visualize the joint density as a solid sitting on the $x$-$y$
plane whose volume must necessarily equal $1$. In this instance, the solid
is a right triangular prism of height $c$, and since the triangular base
has area $1$, the height $c$ must also equal $1$.
For any fixed $x$, the marginal density $f_X(x)$ is given by
$$f_X(x) = \int_{-\infty}^\infty f_{X,Y}(x,y)\,\mathrm dy$$
which is, of course, the area of the cross-section of the
of the joint density solid if we were to slice the solid by
a plane parallel to the $y$-$z$ plane and at distance
$x$ from the $y$-$z$ plane. For $0 < x < 1$, the cross-section
is a rectangle of height $1$ and base extending from $y=-x$
to $y = x$, and so the area is $2x$. For $x\leq 0$ or $x \geq 1$,
the cross-section is $0$. Thus we get
$$f_X(x) = \begin{cases}2x, &0 < x < 1,\\
0, &\text{otherwise.}\end{cases}$$
A similar calculation can be done to obtain the
marginal density $f_Y(y)$. Now, for $0 \leq y < 1$,
the cross-section has base extending from $x = y$ to $1$,
and hence the area is $1-y$, while for $-1 < y \leq 0$,
the cross-section has base extending from $x = -y$ to $1$,
and hence the area is $1+y$. Thus, we have
$$f_Y(y) = \begin{cases}1-y, &0 \leq y < 1,\\
1+y, &-1 < y < 0,\\
0, &\text{otherwise.}\end{cases}$$
As a check on one's work, it is easy to sketch the density
functions and verify that they are nonnegative functions
and the "area under the curve" is $1$, that is, we have
found valid density functions and thus have not made any
glaringly obvious errors in computation.
Finally, to compute $E[Y]$, one can of course use the
standard formula
$$E[Y] = \int_{-\infty}^\infty y f_Y(y)\,\mathrm dy
= \int_{-1}^0 y(1+y)\,\mathrm dy
+ \int_{0}^1 y(1-y)\,\mathrm dy$$
and work out that $E[Y]=0$, but it is also possible to
avoid integration at all by considering that since
$f_Y(y)$ is an even function, the integral
of the odd function $yf_Y(y)$ over the finite interval
$(-1,1)$ must necessarily be $0$. But be sure to remember
that this argument must be used with care if the integral
is over the entire real line. See, for example,
this question
and its answers.