Taking i and j to be the unit vectors in the x and y directions respectively, the question describes a triangle formed by the points (1,0), (0,1) and (0,0), then back to (1,0), and this is the direction of travel along the path.
The line integral is ∮cf(x,y)ds, where ds is an infinitesimal increment in arc length and ds=√(dx2+dy2) by Pythagoras. In this case we have been given differentials in place of f(x,y) and ds has been replaced by dx and dy.
To parameterise this using the parameter t, we can write ds/dt=√((dx/dt)2+(dy/dt)2), which requires us to define functions g(t) and h(t) in place of x and y, and ds/dt=√((dg/dt)2+(dh/dt)2).
We need to consider the three sides of the triangle separately: hypotenuse and two legs. c represents the sum of the three paths, c1, c2 and c3, where c1 is the hypotenuse path, c2 the leg towards the origin and c3 the leg from the origin to the starting point.
(path c1): x+y=1 is the equation of the hypotenuse, so dx+dy=0, that is, dx/dt+dy/dt=0, dy/dt=-dx/dt. The hypotenuse is the line (1,0) to (0,1) which has length √2. Let x=g(t), y=h(t); if x=1-t then x=g(t)=1-t and dg/dt=-1; and y=t, y=h(t)=t and dh/dt=1, making ds/dt=√2, ds=dt√2. Note that when t=0, x=1 and y=0 (the starting point (1,0)); and when t=1, x=0 and y=1 (the endpoint of the hypotenuse (0,1)), providing the limits for t of 0 and 1. Note how the parameterisation is influenced by the direction of the path.
sin(x3)dx-xy+6dy=-sin((1-t)3)dt-(1-t)t+6dt when parameterised for c1. But this expression for f(x,y) contains differentials as well as variables. We can't apply f(g(t),h(t))dt√2 as an integrand. The line integral is essentially the surface area of a "curtain" or "wall" formed by the given triangle as the "floor", and the function f(x,y) (in differential form) as the "ceiling", so f should define a point on the ceiling directly above the point (x,y) on the floor. But this cannot be done if f contains variables as well as differentials.
If we can discover what the integrand is meant to be through some interpretation of what's been given, then the line integral can be evaluated for c1.
(paths c2 and c3): We need this same interpretation to work out the remaining two parts of the path. The hypotenuse brings us to the point (0,1) so parameterisation will be x=0, y=1-t. So dg/dt=dx/dt=0 and dh/dt=dy/dt=-1. Note that when t=0 we are at the point (0,1) and when t=1 we reach the origin (0,0).
The final leg of the path takes us from the origin to (1,0) so here x=t and y=0, dg/dt=dx/dt=1, dh/dt=dy/dt=0, with 0≤t≤1providing the limits.
Now we have the method for finding the 3-piece line integral (if I've understood the question correctly). We still need to work out what f(x,y) actually is.
ANALYSING THE FUNCTION
The problem with the line integral as stated was that the integrand contained dx and dy, but also xy without a differential. The true integrand should have been (I guess):
sin(x3)dx-xydx+6dy or sin(x3)dx-xydy+6dy or, possibly, sin(x3)dx-(xy+6)dy. Then it follows that, by replacing dx with -dt and dy with dt, we get one of three parameterised integrals. Unfortunately we don't know which one applies (if any).
So, for example for c1, there appear to be 3 possible integrals involving only t:
-sin((1-t)3)dt+(t-t2)dt+6dt, -sin((1-t)3)dt-(t-t2)dt+6dt, -sin((1-t)3)dt-(t-t2+6)dt.
For c2 and c3 the integration becomes simpler or rather complicated depending on which terms are going to drop out after substituting for dx and dy.
Until this ambiguity is resolved, the question remains unanswered.