To find the gradient of the curve at the point \((2, 1)\), we need to differentiate the equation implicitly with respect to \(x\).

The given equation is \(2x^2 - 4xy + 3y^2 = 3\).

Differentiate each term:

\(\frac{d}{dx}(2x^2) = 4x\)

\(\frac{d}{dx}(-4xy) = -4y - 4x\frac{dy}{dx}\)

\(\frac{d}{dx}(3y^2) = 6y\frac{dy}{dx}\)

Substitute these into the differentiated equation:

\(4x - 4y - 4x\frac{dy}{dx} + 6y\frac{dy}{dx} = 0\)

Rearrange to solve for \(\frac{dy}{dx}\):

\(-4x\frac{dy}{dx} + 6y\frac{dy}{dx} = 4y - 4x\)

\((6y - 4x)\frac{dy}{dx} = 4y - 4x\)

\(\frac{dy}{dx} = \frac{4y - 4x}{6y - 4x}\)

Substitute \(x = 2\) and \(y = 1\):

\(\frac{dy}{dx} = \frac{4(1) - 4(2)}{6(1) - 4(2)} = \frac{4 - 8}{6 - 8} = \frac{-4}{-2} = 2\)

Thus, the gradient of the curve at the point \((2, 1)\) is 2.