Find a path of a particle I'm stuck with the following question:

Let P be a particle at point $(1,2)$ on the surface $z=x^2y^2$. At $t=0$ the particle is left and moves freely. Find the path that the particle passes during the period of time between $t=0$ and until is stops.

My Try:
First of all, I used GeoGebra to draw the function $f(x,y)=x^2y^2$ and the point $P(1,2,4)$.
I know that the particle will move in the direction of the gradient, hence I calculated the gradient$$\vec{\nabla}f(x,y)=(2xy^2,2x^2y)\underbrace{\Rightarrow}_{x=1,y=2}\vec{\nabla}f=(8,4)$$
Now, I think that the particle will stop when it reaches the origin, but I'm not sure it is true, and if it is true how to explain it.
Also, I don't understand how the calculation of $\vec{\nabla}f(x,y)$ could help me in finding the explicit form of the path.
Please help, thank you.
 A: Assuming the force is the gravity, i.e., $\vec{F}=-g\hat{k}$ and that the surface is frictionless ($\mu=0$), it is obvious that the particle should reach a local minimum of the surface. This minimum $z=0$ is observed whenever $x=0\vee y=0$.
Let $r:(x(t),y(t),z(t))$ the path described by the particle, and $C:(x(t),y(t))$ its projection on the $XY$ plane. To find a parametrization of $C$, we know that the particle will try to follow the direction of maximum decrement on the surface, that is, $-\nabla f(x,y)$ because the gradient is the direction of max increment. This is equivalent to say that the tangent of $C$ is equivalent to $-\nabla f$, that is, 
$$\left(x'(t),y'(t)\right)=-\nabla f(x,y)$$ 
So
$$\begin{array}{rcl}
x'(t)&=&2x(t)y^2(t)\\
y'(t)&=&2x^2(t)y(t)
\end{array}$$
Now, you already calculated the gradient $\nabla f$ at $(1,2)$, so you know all the initial conditions of the system: 
$$\begin{array}{rcl}
(x(0),y(0))&=&(1,2)\\
(x'(0),y'(0))&=&-\nabla f(1,2)=(-8,-4)
\end{array}$$
Now you have to solve the system with initial conditions to get the solution. Good luck!
A: What I would do is the following: Consider the system of differential equations given by  $(x'(t),y'(t))=-\nabla f(x,y)$.  which is the following $x'=-2xy^2$ $y'=-2yx^2$ Since you want the trayectory, you can divide both equations to obtain $y'(x)=x/y$, a first order ODE  you can solve. which solutions are hyperbolas, if $x\neq y$. remember that the particle will stop when $\nabla f=0$, i.e. along the $x$ or $y$  axis
A: The problem here is getting the equations of motion for a particle constrained to a surface. The particle is set free at a certain point on the surface and it moves according (supposedly) to gravity; there is nothing in this setting that tells us that the particle will eventually stop at some point. The behaviour of the particle depends on the particular surface's shape: if it's for instance bowl shaped (for example $z=x^2+y^2$) the particle will "roll" down, go up the other side of the bowl and come back up exactly to the same spot from where it was let free.
This surface is a $2$-dimensional differentiable manifold which has a local (in this lucky case also global) parametrisation $\varphi:\mathbb{R}^2\rightarrow\mathbb{R}^3$, i.e. $\varphi(u,v)=(u,v,u^2v^2)$. We have also a conservative force field, namely a uniform (gravitational) field $G(x,y,z)=(0,0,-g)$. Let's leave friction behind as it complicates the picture quite much.
Such problems are the ideal setting for the Lagrangian formalism. Let $T$ be the kinetic energy of the particle and $U$ its potential energy at a certain point $p=(x,y,z)$ on the surface. Supposing for simplicity that the particle has mass $m=1$, we have
\begin{equation}
T(p)=\frac{1}{2}\|\frac{dp}{dt}\|^2\qquad\text{and}\qquad U(p)=gz
\end{equation}
The Lagrangian of the system is then defined by
\begin{equation}
L:=T-U
\end{equation}
The Lagrangian can also be seen as a function of $u$ and $v$, which are called the generalized coordinates:
\begin{eqnarray}
L(p)&=&L(\varphi(u,v))=\frac{1}{2}\|\frac{d}{dt}\varphi(u,v)\|^2-gu^2v^2=\\
&=&\frac{1}{2}\|(\dot{u},\dot{v},2uv(\dot{u}v+u\dot{v}))\|^2-gu^2v^2=\\
&=&\frac{1}{2}[\dot{u}^2+\dot{v}^2+4u^2v^2(\dot{u}v+u\dot{v})^2]-gu^2v^2
\end{eqnarray}
Then, the equations of motion for the particle in the variables $u$ and $v$ are given by
\begin{equation}
\frac{d}{dt}\left(\frac{\partial L}{\partial\dot{u}}\right)-\frac{\partial L}{\partial u}=0
\qquad\text{and}\qquad
\frac{d}{dt}\left(\frac{\partial L}{\partial\dot{v}}\right)-\frac{\partial L}{\partial v}=0
\end{equation}
These are two differential equations in the variables $u$ and $v$; it seems no easy task solving them, but good luck!
A: Thanks to the given answers I think that I managed to solve the question and would appreciate if someone could check my work.
As suggested we may write that $\displaystyle -\nabla f(x,y)=(\dot{x},\dot{y})$, hence $\displaystyle  \begin{cases}\dot{x}=\frac{dx}{dt}=-2xy^2\\ \dot{y}=\frac{dy}{dt}=-2x^2y\end{cases}$.
Dividing both equations we get $\displaystyle \frac{dy}{dx}=\frac{x}{y}\rightarrow y\cdot dy=x\cdot dx$. Integrating both sides we get the equation $y^2(t)=x^2(t)+C$. Using the initial values $C=3$, hence $y^2(t)=x^2(t)+3$.
Next, I used this relation to write $\displaystyle \frac{dx}{dt}=-2xy^2=-2x(x^2+3)\rightarrow -\frac{dx}{2x(x^2+3)}=dt$.
Integrating both sides $-\displaystyle \int\frac{dx}{2x(x^2+3)}=t+C$. Computing the first integral $$\int\frac{dx}{2x(x^2+3)}=\left[\begin{matrix}x^2=u\\2xdx=du\end{matrix}\right]=\int\frac{du}{4x^2(u+3)}=\frac{1}{4}\int\frac{du}{u(u+3)}=\frac{1}{12}\ln\left(\frac{u}{u+3}\right)+C$$
Hence, $$-\frac{1}{12}\ln\left(\frac{x^2}{x^2+3}\right)=t+C\Rightarrow x(t)=\sqrt{\frac{3}{e^{12(t+C)}-1}}$$ Using the initial values $C=\frac{1}{12}\ln(4)$, thus $\displaystyle x(t)=\sqrt{\frac{3}{4e^{12t}-1}}$.
We can find that $\displaystyle y(t)=x(t)\cdot 2e^{6t}\Rightarrow z(t)=4x^4(t)\cdot e^{12t}$.
We can find that after a short period of time the particle will get to the point $(0,\sqrt{3},0)$ and will stay there.
The only thing I still don't understand is how the initial values of $\nabla f(x,y)$ helps me...
Thanks again for the useful answers!
