If x + 2xy + 3y2 = 1, prove that (x + 3y)3 d2y/dx2 + 2 + 0

$$P(\text{1st Blue}) = \frac{11}{18}$$

After picking one blue ball, there are 17 balls left, with 7 red balls remaining.

$$P(\text{2nd Red | 1st Blue}) = \frac{7}{17}$$ $$P(\text{Blue then Red}) = \frac{11}{18} \times \frac{7}{17} = \frac{77}{306}$$

$$P(\text{1st Red}) = \frac{7}{18}$$

After picking one red ball, there are 17 balls left, with 11 blue balls remaining.

$$P(\text{2nd Blue | 1st Red}) = \frac{11}{17}$$ $$P(\text{Red then Blue}) = \frac{7}{18} \times \frac{11}{17} = \frac{77}{306}$$

$$P(\text{Different Colours}) = P(\text{Blue then Red}) + P(\text{Red then Blue})$$ $$P(\text{Different Colours}) = \frac{77}{306} + \frac{77}{306} = \frac{154}{306}$$

$$\frac{154}{306} = \frac{77}{153}$$

The probability that both balls are of different colours is $\boxed{\frac{77}{153}}$.

Implicit Differentiation Proof:
Given the equation $x^2 + 2xy + 3y^2 = 1$.
We need to prove that $(x+3y)^3 \frac{d^2y}{dx^2} + 2 = 0$.

$$\frac{d}{dx}(x^2) + \frac{d}{dx}(2xy) + \frac{d}{dx}(3y^2) = \frac{d}{dx}(1)$$

Using the product rule for $2xy$ and the chain rule for $3y^2$:

$$2x + (2 \cdot y + 2x \cdot \frac{dy}{dx}) + (3 \cdot 2y \cdot \frac{dy}{dx}) = 0$$ $$2x + 2y + 2x \frac{dy}{dx} + 6y \frac{dy}{dx} = 0$$

$$(2x + 6y) \frac{dy}{dx} = -2x - 2y$$ $$\frac{dy}{dx} = \frac{-2x - 2y}{2x + 6y} = \frac{-2(x+y)}{2(x+3y)}$$ $$\frac{dy}{dx} = -\frac{x+y}{x+3y}$$

Let $u = -(x+y)$ and $v = x+3y$.
Then $\frac{du}{dx} = -(1 + \frac{dy}{dx})$ and $\frac{dv}{dx} = 1 + 3\frac{dy}{dx}$.
The quotient rule is $\frac{d^2y}{dx^2} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$.

$$\frac{d^2y}{dx^2} = \frac{(x+3y) \left( -(1 + \frac{dy}{dx}) \right) - (-(x+y)) \left( 1 + 3\frac{dy}{dx} \right)}{(x+3y)^2}$$ $$\frac{d^2y}{dx^2} = \frac{-(x+3y)(1 + \frac{dy}{dx}) + (x+y)(1 + 3\frac{dy}{dx})}{(x+3y)^2}$$

Numerator:

$$-(x+3y)\left(1 - \frac{x+y}{x+3y}\right) + (x+y)\left(1 - 3\frac{x+y}{x+3y}\right)$$ $$= -(x+3y)\left(\frac{x+3y - (x+y)}{x+3y}\right) + (x+y)\left(\frac{x+3y - 3(x+y)}{x+3y}\right)$$ $$= -(x+3y)\left(\frac{2y}{x+3y}\right) + (x+y)\left(\frac{x+3y - 3x - 3y}{x+3y}\right)$$ $$= -2y + (x+y)\left(\frac{-2x}{x+3y}\right)$$ $$= \frac{-2y(x+3y) - 2x(x+y)}{x+3y}$$ $$= \frac{-2xy - 6y^2 - 2x^2 - 2xy}{x+3y}$$ $$= \frac{-2x^2 - 4xy - 6y^2}{x+3y}$$ $$= \frac{-2(x^2 + 2xy + 3y^2)}{x+3y}$$

From the original equation, we know that $x^2 + 2xy + 3y^2 = 1$.
So, the numerator simplifies to $\frac{-2(1)}{x+3y} = \frac{-2}{x+3y}$.

$$\frac{d^2y}{dx^2} = \frac{\frac{-2}{x+3y}}{(x+3y)^2}$$ $$\frac{d^2y}{dx^2} = \frac{-2}{(x+3y)^3}$$

Multiply both sides by $(x+3y)^3$:

$$(x+3y)^3 \frac{d^2y}{dx^2} = -2$$

Add 2 to both sides:

$$(x+3y)^3 \frac{d^2y}{dx^2} + 2 = 0$$

This proves the statement.

Trigonometry Question:
The question asks to "find the value of $\frac{4 \tan y}{\cos y - \sin y}$" given $0^\circ \le y \le 90^\circ$. This question is incomplete as there is no additional information or equation provided to determine a specific value for $y$ or the expression. Without more context, the expression cannot be simplified to a numerical value.