text: We start by using implicit differentiation:\n\nx² - 2y² = 2\n2x - 4y dy/dx = 0\ndy/dx = x/(2y)\n\nDifferentiating a second time:\n\nd²y/dx² = (2y - 2 dy/dx x)/(4y²)\n\nSubstituting the value we already have for dy/dx:\n\nd²y/dx² = (2y - 2 x/(2y) x)/(4y²)\nd²y/dx² = (2y - x²/y)/(4y²)\n\nSolving the original equation for x²:\n\nx² - 2y² = 2\nx²=2+2y²\n\nAnd substituting:\n\nd²y/dx² = (2y - (2+2y²)/y)/(4y²)\nd²y/dx² = (2y - 2/y - 2y)/(4y²)\nd²y/dx² =  (-2/y)/(4y²)\nd²y/dx² =  -1/(2y³)\n\nd²y/dx² = -1/2 * y^(-3)\n\nso a=-1/2 and b=-3