<s>[INST] How does one project the gradient at a point on a surface into a plane? I am studying Multivariable Calculus and have come to the following excerpt in my book:

I can see clearly how they get from the given function to 
$$ y'(x)\ =\ \frac{3y}{x} $$
And understand that this slope passes through the given point. The following line leaves me totally lost, however:

You can verify that the solution to this differential equation is $$ y\ =\ \frac{4x^3}{27} $$ and the projection of the path of steepest descent in the xy-plane is the curve $$ y\ =\ \frac{4x^3}{27} $$

How did they get from any of the given information--the initial equation, its gradient, the slope of the gradient, etc.--to the above equation for y? Furthermore, how am I to find the projection of the gradient in the xy-plane?
 [/INST] Thanks to user @Moo for helping out in the comments on the original post.
The solution is simple: you know the first derivative of the line and a point it goes through. If you take the first, indefinite integral of your derivative--you get the original line, plus some constant C. Plug your point into this line, solve for C, and--voila! You have your answer.
\begin{align*}
 y'(x) &= 3\frac{y}{x} \\
 \frac{dy}{dx}\ &= 3\frac{y}{x} \\
\frac{1}{y} dy\ &= 3\frac{1}{x}\ dx \\
\int\frac{1}{y} dy\ &= 3\int\frac{1}{x}\,dx \\
\ln(y) + C_y\ &= 3\ln(x) +C_x \\
\ln(y) &= 3\ \ln(x) + C \\
\ln(y) - 3 \ln(x)\ &= C \\
\ln(y) - \ln\left(x^3\right)\ &= C \\
\ln \left(\frac{y}{x^3}\right) &= C \\
\ln \left(\frac{4}{3^3}\right) &=\ C \\
\ln \left(\frac{4}{27}\right) &=\ C \\
\ln(y) &= 3 \ln(x) + \ln \left(\frac{4}{27}\right) \\
\ln(y) &= \ln(x^3) + \ln \left(\frac{4}{27}\right) \\
\ln(y) &= \ln\left(\frac{4x^3}{27}\right) \\
y &= \frac{4x^3}{27} \\
\end{align*}
</s>