Solve for differential equation $dy/dx=dx/dy$ My approach is:
$$\int\int dydy=\int\int dxdx$$
$$(y+a)^2=(x+b)^2$$
$$y=\pm x+c$$
I am not sure if I made mistakes, though, especially the first and second steps. Could you please check it? Any answers will be appreciated.

Are there anything wrong with:
$$\int\int dydy=\int\int dxdx$$
$$\int(y+c_1)dy=\int(x+c_2)dx$$
$$(y+a)^2=(x+b)^2+c\ \  ???$$
This seems inconsistent with the correct  answer.
 A: Your result is correct.
Another approach :
$$\frac{dy}{dx}=\frac{dx}{dy}\quad\to\quad \left(\frac{dy}{dx} \right)^2=1\quad\to\quad \frac{dy}{dx}=\pm 1 \quad\to\quad y=\pm x+c$$
A: Your approach, even though it leads to the correct result, is black magic to me. It is true that for ODEs of the type$$y'=f(x)g(y)$$
we use the method of separating variables:
$$\int_{y_0}^y{1\over g(y')}\>dy'=\int_{x_0}^x f(x')\>dx'\ ,$$
because  long ago we have seen the (not so simple) proof that this is o.k. On the other hand, I don't know what $\int\int dy\>dy$ means (shouldn't there be two constants of integration in the end?), nor do I have the slightest trust that your playing with the symbols (even if meaningful) will lead to a correct result.
Your differential equation is saying no more and no less than $y'={1\over y'}$, and then should be solved along the lines of JJacquelin's answer.
A: Just for curiosity :
In order to avoid supernumerary solutions, one have to use definite integrals instead of indefinite. It becomes more complicated : 
$$\int_{y_2=-a}^{y_2=y}\left(\int_{y_1=-a}^{y_1=y_2}dy_1 \right)dy_2 = \int_{x_2=-b}^{x_2=x}\left(\int_{x_1=-b}^{x_1=x_2}dx_1 \right)dx_2$$
with arbitrary $a$ and $b$.
Note : For the term at left be equal to the term at right, the starting point for integration must be the same on both sides, say $(-a,-b)$. These notation is chosen to be consistent with the notation of the previous answers. This explains why the lower bounds are common for each double integral.
$$\int_{y_2=-a}^{y_2=y}\left(y_2+a \right)dy_2 = \int_{x_2=-b}^{x_2=x}\left(x_2+b \right)dx_2$$
$$\left[\frac{y_2^2}{2}+ay_2+c_1\right]_{y_2=-a}^{y_2=y} = \left[\frac{x_2^2}{2}+bx_2+c_2\right]_{x_2=-b}^{x_2=x}$$
$$\left(\frac{y^2}{2}+ay+c_1\right)-\left(\frac{(-a)^2}{2}+a(-a)+c_1\right) = \left(\frac{x^2}{2}+bx+c_2\right)-\left(\frac{(-b)^2}{2}+b(-b)+c_2\right)$$
$$\frac{y^2}{2}+ay+\frac{a^2}{2} =\frac{x^2}{2}+bx+\frac{b^2}{2}$$
$$(y+a)^2=(x+b)^2$$
$$y+a=\pm(x+b) \quad\to\quad y=\pm x+c$$
Obviously this method is not recommended compared to the method with $y'^2=1$.
A: $$\frac{dy}{dx}=\frac{dx}{dy}\implies\int\frac{dy}{dx}dx=\int\frac{dx}{dy}dx=\frac{d}{dy}\int x dx$$
Thus,
$$\int \int \frac{dy}{dx} dx dy = \int\frac{d}{dy}\int x dx dy$$
Looking at the l.h.s.,
$$\int \int \frac{dy}{dx} dx dy=\int (y + a) dy=\frac{(y+a)^2}{2}+c_1$$
On the r.h.s.,
$$\int\frac{d}{dy}\int x dx dy= \int\frac{d}{dy}\left(\frac{x^2}{2}+c_2\right)dy=\frac{x^2}{2}+c_2$$
Thus,
$$\frac{(y+a)^2}{2}+c_1=\frac{x^2}{2}+c_2$$
Or,
$$(y+a)^2=x^2+b$$
Now, recognize that 
$$\frac{d}{dy}(y+a)^2 = 2x \implies \frac{dy}{dx}=\frac{x}{y+a}$$
Similarly,
$$2(y+a)=2x\frac{dx}{dy}\implies\frac{dx}{dy}=\frac{y+a}{x}$$
In order to satisfy the original equation, $\frac{dy}{dx}=\frac{dx}{dy}$ we conclude that $b=0$. Thus, 
$$(y+a)^2 = x^2$$
A: Alternatively:
$$\frac{dy}{dx}=\frac{dx}{dy}=k \Rightarrow \begin{cases} dy=kdx \\ \frac{dx}{k}=dy \end{cases} \Rightarrow \begin{cases} y=kx+C \\ y=\frac{1}{k}x+C\end{cases} \Rightarrow k=\frac{1}{k} \Rightarrow k=\pm 1 \Rightarrow y=\pm x+C.$$
