a. [3 marks]

Given: \( e^{x+y} = x^2 + y^2 \).
Differentiate implicitly: Left side: \(\frac{d}{dx} [e^{x+y}] = e^{x+y} (1 + \frac{dy}{dx})\) (M1).
Right side: \(\frac{d}{dx} [x^2 + y^2] = 2x + 2y \frac{dy}{dx}\) (M1).
Equate: \( e^{x+y} (1 + \frac{dy}{dx}) = 2x + 2y \frac{dy}{dx} \).
Rearrange: \( e^{x+y} \frac{dy}{dx} – 2y \frac{dy}{dx} = 2x – e^{x+y} \implies \frac{dy}{dx} (e^{x+y} – 2y) = 2x – e^{x+y} \implies \frac{dy}{dx} = \frac{2x – e^{x+y}}{e^{x+y} – 2y} \) (A1).
Answer: \(\frac{dy}{dx} = \frac{2x – e^{x+y}}{e^{x+y} – 2y}\).

b. [3 + 3 marks]

(i) Horizontal tangents at \( P \) and \( Q \): \(\frac{dy}{dx} = 0 \implies 2x – e^{x+y} = 0 \implies e^{x+y} = 2x \implies x + y = \ln(2x)\) (M1).
Substitute \( y = \ln(2x) – x \) into \( e^{x+y} = x^2 + y^2 \): \( 2x = x^2 + (\ln(2x) – x)^2 \) (M1).
Expand: \( (\ln(2x) – x)^2 = (\ln(2x))^2 – 2x \ln(2x) + x^2 \).
Thus: \( 2x = x^2 + (\ln(2x))^2 – 2x \ln(2x) + x^2 \implies 2x^2 + (\ln(2x))^2 – 2x \ln(2x) – 2x = 0 \) (A1).
Answer: \( 2x^2 + (\ln(2x))^2 – 2x \ln(2x) – 2x = 0 \).
(ii) Solve numerically: \( 2x^2 + (\ln(2x))^2 – 2x \ln(2x) – 2x = 0 \).
Approximate solutions: \( x \approx 0.331077, 1.84273 \) (M1).
Find \( y \): \( y = \ln(2x) – x \).
For \( x \approx 0.331077 \): \( y \approx \ln(2 \cdot 0.331077) – 0.331077 \approx -0.743332 \).
For \( x \approx 1.84273 \): \( y \approx \ln(2 \cdot 1.84273) – 1.84273 \approx -0.538335 \) (A1).
Answer: Coordinates of \( P \): \( (0.331, -0.743) \), \( Q \): \( (1.84, -0.538) \) (A1).

c. [2 marks]

Curve is symmetric about \( y = x \). Vertical tangents occur where horizontal tangents’ coordinates are swapped (M1).
From part (b)(ii): \( P (0.331, -0.743) \), \( Q (1.84, -0.538) \).
Swap coordinates: Vertical tangents at \( (-0.743, 0.331) \), \( (-0.538, 1.84) \) (A1).
Answer: \( (-0.743, 0.331) \), \( (-0.538, 1.84) \).

d. [3 marks]

Method 1:
Set gradient: \(\frac{2x – e^{x+y}}{e^{x+y} – 2y} = -1 \implies 2x – e^{x+y} = -(e^{x+y} – 2y) \implies 2x = 2y \implies y = x\) (M1).
Substitute \( y = x \) into \( e^{x+y} = x^2 + y^2 \): \( e^{2x} = 2x^2 \) (M1).
Solve numerically: \( x \approx -0.451 \) (A1).
Thus: \( y = x \approx -0.451 \).
Answer: \( (-0.451, -0.451) \).
Method 2:
Recognize point lies on line of symmetry \( y = x \) (since gradient \(-1\) is perpendicular to line \( y = x \)) (M1).
Substitute \( y = x \) into \( e^{2x} = 2x^2 \).
Solve numerically: \( x \approx -0.451 \) (A1).
Thus: \( y = x \approx -0.451 \) (A1).
Answer: \( (-0.451, -0.451) \).