The question is correctly stated but you are confusing yourself about what is required. $X$ and $Y$ are independent random variables, the
meaning of which (not the math) is that knowing the value of one of them
tells you nothing that you don't already know about the other. The math
says that
$$\begin{align}
f_{X,Y}(x,y) &= f_X(x)f_Y(y) &\text{by independence}\\
&= \exp(-x)\exp(-y), &\qquad 0 \leq x < \infty, ~ 0 \leq y < \infty,\\
&= \exp(-x-y), &\qquad 0 \leq x < \infty, ~ 0 \leq y < \infty,
\end{align}$$
Now, the random point $(X,Y)$ can be anywhere in the first quadrant
of the $x$-$y$ plane. If you are told that $Y$ took on the value
$\alpha$, then you know that the random point lies somewhere on
the horizontal line at height $\alpha$ above the $x$ axis, but
you get no information about $X$. In particular, the conditional
density $f_{X\mid Y = \alpha}(x\mid Y = \alpha)$ of $X$ given $Y=\alpha$
is the same as the unconditional density $f_X(x)$. If you want
an explicit working out of this answer that crosses all i's and
dots all t's, just use the standard formula
$$f_{X\mid Y = \alpha}(x\mid Y = \alpha)
=\frac{f_{X,Y}(x,\alpha)}{f_Y(\alpha)}
= \frac{\exp(-x-\alpha)}{\exp(-\alpha)} = \exp(-x), ~~~0 < x \leq \infty.$$
The way to think about the conditional density formula
used above (and not just for the independent case here)
is that the shape of the conditional density of $X$
given $Y = \alpha$ is just the curve $f_{X,Y}(x,\alpha)$.
This is not the actual conditional density because the
area under the curve is not necessarily $1$ as it must be
for all density functions. But, given any curve $g(x)$
such that $g(x) \geq 0$ for all $x$ and the area under
the curve is finite, the function
$$\frac{g(x)}{\int_{-\infty}^\infty g(x)\,\mathrm dx}$$
is a valid density function. Note that the denominator
above is just the (finite) area under the curve $g(x)$.
For our application to conditional density calculations,
we need to divide $f_{X,Y}(x,\alpha)$ by the area under
$f_{X,Y}(x,\alpha)$, but this area is
$$\int_{-\infty}^\infty f_{X,Y}(x,\alpha)\,\mathrm dx$$
which I hope you will recognize as just the calculation
that needs to be done to find the value of the
marginal density $f_Y(y)$ at $y = \alpha$, that is,
the above integral gives us $f_Y(\alpha)$, and the
ratio $f_{X,Y}(x,\alpha)/f_Y(\alpha)$ is thus the
conditional density function $f_{X\mid Y = \alpha}(x\mid Y = \alpha)$
Turning to the question asked here, it is slightly trickier
in that the
information given is that the random point $(X,Y)$ lies on
the straight line $x+y=c$, but the answer is a lot
simpler. Now, since the point must
be in the first quadrant, you know it lies on the line
segment with end points $(0,c)$ and $(c,0)$. Now you
do get some information about $X$; you know immediately
that $X$ can take on values only in the interval $[0,c]$.
The conditional density of $X$ given $X+Y=c$ is just the
function $f_{X,Y}(x,y)$ along the curve $x+y=c$. But since
$f_{X,Y}(x,y) = \exp(-x-y) = \exp(-c)$ along this curve,
the conditional density of $X$ given $X+Y=c$ has constant
value for $0 \leq x \leq c$, that is,
The conditional density of $X$ given $X+Y=c$ is a
uniform density on $[0,c]$.
In fact, a little thought will show that the conditional
density of $Y$ given $X+Y=c$ is also a
uniform density on $[0,c]$. But of course, $X$ and $Y$
are not conditionally independent given $X+Y=c$ since
each is constrained to be $c$ minus the other.