Answer
work done $\int_C \textbf F \cdot \textbf{dr} = -\frac{43}{105}$ units
Work Step by Step
The path is given by
$C: x=\cos^3t, y = \sin^3t$ from (1,0) to (0,1)
This path can be written as
$ \textbf r = x\textbf i + y\textbf j + z \textbf k\\
= \cos^3 t \textbf i + \sin^3t \textbf j, \text{for } 0\le t \le \pi/2\\
\therefore \textbf r' = -3\cos^2 t\sin t \textbf i + 3\sin^2 t \cos t \textbf j$
Given force field $\textbf F = x^2 \textbf i -xy \textbf j\\
=\cos^6 t \textbf i - \cos^3 t \sin^3t \textbf j$
work done = $\int_C \textbf F \cdot d\textbf r = \int _{t=0}^{\pi /2 }\textbf F \cdot \textbf r' dt\\
= \int _{0}^{\pi /2 }(\cos^6 t \textbf i - \cos^3 t \sin^3t \textbf j)\cdot (-3\cos^2 t\sin t \textbf i + 3\sin^2 t \cos t \textbf j) dt\\
= \int _{0}^{\pi /2 } [-3\cos^8 t .\sin t - 3 \cos^4t.\sin^5t]dt\\
=\int _{0}^{\pi /2 } [-3\cos^8 t .\sin t - 3 \cos^4t. (1-\cos^2t)^2 \sin t]dt\\
= \int _{0}^{\pi /2 } [-6\cos^8t.\sin t+6\cos^6t.\sin t-3\cos^4t.\sin t]dt\\
=[\frac{6}{9}\cos^9t-\frac{6}{7}\cos^7t + \frac{3}{5}\cos^5t]_0^{\pi/2}\\
= -\frac{43}{105}$
workdone = $-\frac{43}{105}$ units
