First thing to notice is $y = \ln\left(\mathrm{e}^{y}\right)$
$$
\ln(y') = x - \ln\left(\mathrm{e}^{y}\right) -\mathrm{e}^{y}
$$
then we have
$$
\ln\left(y'\mathrm{e}^{y}\right) = x - \mathrm{e}^{y}
$$
using the sub $v = \mathrm{e}^{y}$ leads to
$$
\ln(v') = x- v
$$
thus
$$
v' = \mathrm{e}^{x}\mathrm{e}^{-v}
$$
hence
$$
\mathrm{e}^{v} = \mathrm{e}^{x}+C\implies v = \ln\left(\mathrm{e}^{x}+C\right)
$$
and subbing in for y
$$
y(x) = \ln\left[\ln\left(\mathrm{e}^{x}+C\right)\right]
$$
now we have $y(1) = 0$ which means
$$
y(1) = 0 = \ln\left[\ln\left(\mathrm{e}+C\right)\right]
$$
therefore $C = 0$ so the solution is actually
$$
y(x) = \ln(x)
$$
you could check that the solution you found does not hold for the original equation.
This question is a bit of a nightmare for an entrance exam. Anyway, first observe that $$d(x^2 + y^2) = 2xdx + 2ydy$$ This suggests that using a variable $u = x^2 + y^2$ is useful, as we also have the RHS (right hand side)
$$RHS = \sqrt{\frac{a^2 - u}{u}}$$
The denominator on the LHS (left hand side) is slightly tricker; the minus sign suggests a derivative of $1/x$. Let's try $v = y/x$, then $$dv = \frac{1}{x} dy - \frac{y}{x^2} dx$$
Hence
$x^2 dv = xdy - ydx$ and we can write the LHS
$$LHS = \frac{x dx + y dy}{x dy - y dx} = \frac{1}{2x^2} \frac{du}{dv}$$
If we can write the $x^2$ terms of $u$ and $v$ we will have an ODE in just those variables: $$ \frac{1}{v^2 + 1} = \frac{x^2}{x^2 + y^2} = \frac{x^2}{u} \ \ \hbox{ hence } \ x^2 = \frac{u}{v^2 + 1}$$ Thus we can write
$$LHS = \frac{v^2 + 1}{2u} \frac{du}{dv} = RHS = \sqrt{\frac{a^2 - u}{u}}$$
or
$$\frac{du}{dv} = 2\sqrt{u(a^2 - u)} . \frac{1}{v^2 + 1}$$
This equation is separable
$$\int \frac{du}{\sqrt{u(a^2 - u)}} = 2\int \frac{dv}{v^2 + 1}$$
...after a bit of work,
$$\arctan\left( \frac{\sqrt{u}}{\sqrt{a^2 - u}} \right) = \arctan v + C$$
or back in the original variables:
$$\arctan\left( \frac{\sqrt{x^2 + y^2}}{\sqrt{a^2 - x^2 - y^2}} \right) = \arctan\left(\frac{y}{x} \right) + C$$
Best Answer
The solution (below) is obtained on the form of an implicit equation : $$(y^2-x^2+1)^5-c\:(y^2+x^2-3)=0$$ where $c$ is a constant.
Since it's a quintic polynomial equation, $y^2$ cannot be expressed as a function of $x^2$ with a finite number of elementary functions.
Theoretically, $y(x)$ could be expressed on a closed form thanks to special functions (in terms of Jacobi theta functions), but this would be very arduous : http://mathworld.wolfram.com/QuinticEquation.html