Orthogonal trajectories of the family of hyperbolas I was looking for the orthogonal trajectories of the family of hyperbolas $x^2 - y^2 = ay. $ This is how I got the orthogonal trajectories:
Doing the implicit differentiation, we get $$2x - 2yy' = ay'$$ Rearranging it, it looks like this: $$\frac{dy}{dx} = \frac{2x}{a+2y}$$
Getting rid of $a$ in the above equation, it looks like this:$$\frac{dy}{dx} = \frac{2x}{\frac{x^2 - y^2}{y}+2y}$$
Rearranging it, it looks like this: $$\frac{dy}{dx} = \frac{2xy}{x^2 + y^2}$$
Since the slope of the curve of an orthogonal trajectories is the negative reciprocal of the slope of the given family of curve above, the slope of the curve of an orthogonal trajectories is then: $$\frac{dy}{dx} = \frac{x^2 + y^2}{-2xy}$$
Rearranging it to see the differential equation that would describe the orthogonal trajectories, it looks like this:$$-2xydy = (x^2 + y^2)dx$$
Rearranging it, it looks like this:$$-x^2 dx - 2xy dy = y^2 dx$$ or $$x^2 dx  + 2xy dy = -(y^2 dx)$$
I noticed that the differential equation has an integrable combinations, so it looks like this:$$d(x^2y) = d \left(\frac{y^3}{-3} \right) $$ 
Then doing this:$$\int d(x^2y) = \int d \left(\frac{y^3}{-3} \right) $$
The equation of the curve of the orthogonal trajectories that passes through the family of hyperbolas $x^2 - y^2 = ay $ is $$x^2y + c = \frac{y^3}{-3} + c$$ or $$y^3 - 3x^2y = c$$
The funny thing is, the answer given in the book is $x^3 + 3xy^2 = c.$ Where did I messed up?
 A: Let's try those "integrable combinations" again, using the rules of differentiation:
\begin{align}\newcommand{d}{\mathrm d}
 \d\!\left(x^2y\right) &= y\,\d\!\left(x^2\right) + x^2\,\d(y)   && \text{(product rule)} \\
          &= 2xy\,\d x + x^2\,\d y     && \text{(power rule)}\\[6pt]
 \d\!\left(\frac{y^3}{-3} \right) &= -\frac13\,\d\!\left(y^3\right)
   && \text{(constant factor)} \\
   &= -\frac13\left(3y^2\,\d y\right)    && \text{(power rule)} \\
   &= -y^2\,\d y
\end{align}
So the equation $\d\!\left(x^2y\right) = \d\!\left(\frac{y^3}{-3} \right)$
is equivalent to $x^2\,\d y + 2xy\,\d x = -y^2\,\d y,$
whereas you were looking for something equivalent to
$x^2\,\d x + 2xy\,\d y = -y^2\,\d x.$
In other words, you accidentally swapped $\d x$ with $\d y.$

Note that this answer does not say there is anything wrong with the equation
$$x^2\,\d x + 2xy\,\d y = -y^2\,\d x. \tag1$$
In fact, that is a correct equation, and if you convert this equation (or something equivalent) into differentials you can solve the problem.
The error is not in Equation $(1),$
but rather in the substitution of $\d\!\left(x^2y\right)$ for 
$x^2\,\d x + 2xy\,\d y$ (incorrect, because these two things are not equal)
and the substitution of $\d\!\left(y^3/(-3)\right)$ for
$-y^2\,\d x$ (also incorrect).
Rather than correcting an "error" in Equation $(1),$ you need to correct the substitutions of the differentials.
But also note that Equation $(1)$ is merely one of many equivalent equations, and there was no particular reason you had to choose an equation in which $-(y^2 dx)$ was alone on right-hand side.
Instead of writing Equation $(1)$ 
you could equivalently have written $$y^2 dx  + 2xy dy = -(x^2 dx).\tag2$$
And now you may notice that the two sides of Equation $(2)$  are just like the differentials worked out in the first part of this answer,
except that all the $x$s have been replaced by $y$s and vice versa.
This suggests replacing $x$ with $y$ and $y$ with $x$ in the differentials:
$$
\d(y^2x) \quad\text{and}\quad \d\!\left(\frac{x^3}{-3} \right).
$$
Try working out those differentials and see if they allow you to make correct substitutions in Equation $(2).$
