(a) Differentiate the equation \(3x^2 + 4xy + 3y^2 = 5\) implicitly with respect to \(x\):

\(\frac{d}{dx}(3x^2) = 6x\),

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

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

Set the derivative equal to zero:

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

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

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

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

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

\(\frac{dy}{dx} = -\frac{3x + 2y}{2x + 3y}\).

(b) The slope of the tangent is \(-2\) (parallel to \(y + 2x = 0\)).

Set \(\frac{dy}{dx} = -2\):

\(-\frac{3x + 2y}{2x + 3y} = -2\),

\(3x + 2y = 4x + 6y\),

\(x = -4y\) or \(y = -\frac{x}{4}\).

Substitute \(x = -4y\) into the curve equation:

\(3(-4y)^2 + 4(-4y)y + 3y^2 = 5\),

\(48y^2 - 16y^2 + 3y^2 = 5\),

\(35y^2 = 5\),

\(y^2 = \frac{1}{7}\),

\(y = \pm \frac{1}{\sqrt{7}}\).

For \(y = \frac{1}{\sqrt{7}}\), \(x = -4\left(\frac{1}{\sqrt{7}}\right) = -\frac{4}{\sqrt{7}}\).

For \(y = -\frac{1}{\sqrt{7}}\), \(x = 4\left(\frac{1}{\sqrt{7}}\right) = \frac{4}{\sqrt{7}}\).

The coordinates are \(\left( \frac{4}{\sqrt{7}}, \frac{1}{\sqrt{7}} \right)\) and \(\left( -\frac{4}{\sqrt{7}}, -\frac{1}{\sqrt{7}} \right)\).