A two-dimensional incompressible frictionless flow field is given by   If ρ is the density of the fluid, the expression for pressure gradient vector at any point in the flow field is given as 

The given two-dimensional velocity field is:
$\overrightarrow{u}=x\hat{i}-y\hat{j}$
From this velocity vector, the velocity components in the $x$ and $y$ directions are:
$u=x$
$v=-y$

For a two-dimensional, incompressible, and frictionless flow, the Euler equations of motion in the $x$ and $y$ directions (neglecting body forces) are:
$-\frac{1}{\rho }\frac{\partial P}{\partial x}=u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}$
$-\frac{1}{\rho }\frac{\partial P}{\partial y}=u\frac{\partial v}{\partial x}+v\frac{\partial v}{\partial y}$

First, let's calculate the partial derivatives of the velocity components:
$\frac{\partial u}{\partial x}=1$, $\frac{\partial u}{\partial y}=0$
$\frac{\partial v}{\partial x}=0$, $\frac{\partial v}{\partial y}=-1$

Substitute these values into the acceleration component equations:
For the $x$-direction:
$a_x=u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}=\left(x\right)\left(1\right)+(-y)\left(0\right)=x$
Thus, the pressure gradient in the $x$-direction is:
$\frac{\partial P}{\partial x}=-\rho x$

For the $y$-direction:
$a_y=u\frac{\partial v}{\partial x}+v\frac{\partial v}{\partial y}=\left(x\right)\left(0\right)+(-y)(-1)=y$
Thus, the pressure gradient in the $y$-direction is:
$\frac{\partial P}{\partial y}=-\rho y$

The pressure gradient vector $\nabla P$ is given by:
$\nabla P=\frac{\partial P}{\partial x}\hat{i}+\frac{\partial P}{\partial y}\hat{j}$
Substituting the derived gradients:
$\nabla P=-\rho x\hat{i}-\rho y\hat{j}=-\rho \left(x\hat{i}+y\hat{j}\right)$

This matches the expression shown in the second option.