If $F=\vec{\triangledown}W$, then:
$F_x=\dfrac{\partial W}{\partial x}$ and $F_y=\dfrac{\partial W}{\partial y}$
Therefore:
$$W=\int F_x dx = \int F_y dy$$
Now if we look at it in another way, we get a different result:
$$W= \int \left( \dfrac{\partial W}{\partial x} dx+ \dfrac{\partial W}{\partial y}dy \right) $$ (by multivariable chain rule)
$$=\int \left( F_x dx+F_y dy \right)$$ $$=\int F_x dx+ \int F_y dy$$
This time I am getting a different result for $W$. Why am I getting contradictory results? Please explain.
$W = \int F_x \; dx$ just says that $W$ is an antiderivative of $F_x$ with respect to $x$, i.e. that $\dfrac{\partial W}{\partial x} = F_x$. Using the Fundamental Theorem of Calculus, that translates to $$ W(x_1, y) - W(x_0, y) = \int_{x_0}^{x_1} F_x(x,y)\; dx$$ Note that this is all with the same $y$.
If you want to relate $W$ at two points with different $y$'s as well as different $x$'s, then you can use
$$\eqalign{ W(x_1, y_1) - W(x_0, y_0) &= W(x_1, y_1) - W(x_1, y_0) + W(x_1, y_0) - W(x_0, y_0) \cr &= \int_{y_0}^{y_1} F_y(x_1, y)\; dy + \int_{x_0}^{x_1} F_x(x, y_0)\; dx} $$ or more generally, for any (piecewise smooth) path $C$ from $(x_0, y_0)$ to $(x_1, y_1)$, the path integral
$$ W(x_1, y_1) - W(x_0, y_0) = \int_C \left( F_x(x,y)\; dx + F_y(x,y)\; dy\right)$$