When I'm given a velocity vector (they call it "velocity field") that's $a\vec{i} + xy\vec{j},$ its easy to integrate both of them with respect to time separately.
Integrating $a\vec{i}$ becomes $x - x_0 = a(t-t_0)$ and then $\vec{i}$ solve for $x,$ substitute into the other vector to eliminate the $x,$ and integrate the $\vec{j}$ vector.
I end up with an equation for each vector, no problem, I can substitute my given times in and get the new position of the particle - the first one gives me my $x$ position and the second one my $y$ position.
Where my problem lies is I don't know how to do those steps when given a more complicated vector like $xy^2\vec{i} + x^2y\vec{j}.$
If I integrate the left side first with respect to time and solve for $x,$ first, I'm not sure if I'm supposed to treat $y$ as a constant, and second, when I solve for $x$ and substitute into the $\vec{j}$ vector I end up with an insane equation that is extremely difficult, if not impossible for me to integrate.
Is there an easier method that I don't know about so that I can integrate my original vector so that given two times I can determine its new $x,y$ position?
You are not supposed to treat $y$ as a constant when you integrate the $\vec{i}$ component with respect to $t$. Both $x$ and $y$ are assumed to be functions of time.
What I'm understanding is that you're treating the vector field $$\vec{F}(x,y)=\Big<xy^2,x^2y\Big>$$ as the flow-velocity of some static fluid, and you are attempting to solve the system of differential equations $$x'(t)=xy^2$$ $$y'(t)=x^2y$$
Note how $$\frac{dy}{dx}=\frac{y'(t)}{x'(t)}=\frac{x}{y}$$
You can solve this first order ODE and see how a solution $(x(t),y(t))$ to your system must satisfy $$(y(t))^2 - (x(t))^2=C$$ where $C$ is an arbitrary constant depending on your initial condition. Using this fact, you can write $$\frac{dx}{dt}=x(x^2+C)$$ and solve for $x$.
There are many well$-$established techniques for solving, or approximating solutions to, systems of non$-$linear differential equations. You just have to look them up!