Find an equation of the curve that passes through the point $(0, 6)$ and whose slope at $(x, y)$ is $\frac{x}{y}$. Book wasn't helpful. I am using James Stewarts Early Transcendentals Calculus, and Section 9.3 (which is where this problem comes from) doesn't seem to have anything remotely similar to the problem I am facing. No examples, nothing
I hate to just dump a homework problem off, but I really don't know where to begin. I thought "find the tangent line" at first, but it doesn't mention tangent anywhere, and I only thought of that because of previous math classes where I had to find the equation of a tangent line.
Reprinted problem below:

Find an equation of the curve that passes through the point
   $(0, 6)$ and whose slope at $(x, y)$ is $\frac{x}{y}$.

 A: The slope at $(x,y)$ is another way of saying the derivative of the curve, so the question can be read as find the equation of a line that satisfies
\begin{equation}
\frac{dy}{dx} = \frac{x}{y},
\end{equation}
and that passes through the point $(x,y) = (0,6)$.
Hint: this is a first order ODE that can be solved and will have one constant of integration that can be set such that the solution passes through the desired point. 
A: You can use WolframAlpha to check your result: (link)
It recognizes the differential equation is separable:
$$
y'(x) \, y(x) = x
$$
And you can solve this by integrating both sides.
$$
y \, \frac{dy}{dx} = x \iff \\
\int\limits_{y(x_0)}^{y(x)} y(\xi) \, dy = \int\limits_{x_0}^x \xi \, d\xi \iff \\
\frac{1}{2} \left( y(x)^2 - y(x_0)^2 \right) = x - x_0
$$
In our case $(x_0, y(x_0)) = (0,6)$, so we get:
$$
\frac{1}{2} \left(y(x)^2 - 36 \right) = x \iff \\
y(x) = \sqrt{x+ 36}
$$
A: It is variable separable differential equation.Integrating,
$$ \frac{dy}{dx} = \frac{x}{y}, \, y {dy}-x {dx} = 0,\, y^2/2 -x^2/2 = c$$
Plug in BC:
$$ 6^2 /2-0 = c = 18,\,  y^2-x^2= 36. $$
There is  no calculus law about y/x, any relation that agrees with the differentials seen as connecting separate terms is ok.
A: Assume that $y$ is never $0.$ We have $$y'=x/y\implies yy'=x\implies (y^2)'=2 y y'=2 x.$$ So $y^2$ is an anti-derivative of the function $g(x)=2 x.$ So $y^2=x^2+K$  for some constant $K.$ When $x=0$ we have $36=6^2=y^2=x^2+K=K.$ So $y=\sqrt {x^2+36}.\quad$(We take the positive square root because $y>0$ when $x=0.$)    
