Solution:
\( \displaylines{\text{The gradient is }\nabla f\left(x,y\right)=\lt12x+4y,4x-6y\gt\\ \text{Evaluating at the given point, }P\text{, gives }\lt68,30\gt\\ \text{The magnitude is }\sqrt{68^2+30^2}\approx74.3\implies\lt0.915,0.404\gt\\ \\ \text{A vector in the direction of no change would be perpendicular to the steepest ascent vector.}\\ \text{We can solve}\lt68,30\gt\cdot\lt a,b\gt=0\text{ for }a,b\text{ to find such a vector.}\\ 68a+30b=0\implies b=-\frac{68}{30}a\\ \lt30,-68\gt\text{ is the direction (or its negation).}} \)