Implicit Differentiation. Example Problem 3 For what values of x does the implicit equation x cubed plus ycubed = 3xy have a horizontal tangent?Again, we're taking the derivative. Specifically, a horizontaltangent indicates that dy dx = 0, whereas a vertical tangentindicates that dx dy = 0.Let's begin by finding the derivative of x cubed.The derivative ofx cubed is 3x squared.For y cubed, we get the same result.The derivative of y is dy dx plus the derivative of 3x is3.Keeping only the y term, we find that dy dx equals 3y.We will isolate dy dx.We are going to move all of the terms with dy dx to the left sideof the equation, leaving a 3y minus 3x squared term on the rightside. We can then factor out a dy dx from this expression. 3yminus 3x squared.We are striving to find the value of dy/dx that produces ahorizontal tangent, so we need a numerator of 0. We also need thedenominator not to be 0 because if you get 0 over 0, it's notnecessarily a horizontal tangent or a vertical tangent.The graph is indeterminate in the first quadrant.As long as the numerator of a fraction is 0 and the denominator isnonzero, there will be no solution to that equation when x = y =0. If x = 1 or y = 1, then there will not be a solution.Let y equal x squared divided by 3, so that 3y squared equals 3xsquared. Letting y equal x squared gives us a value for y that wecan substitute into the equation to find x.So, as long as y is equal to x squared, we will have horizontaltangents. However, we must make sure that our solution is a pointon the original curve; therefore, we will substitute back into theoriginal equation and solve for y.x cubed plus y cubed is equal to 3xy.We were told that y equals x squared, so everywhere we see y inthe equation, we can replace it with x squared. So the equationbecomes x cubed plus x to the sixth is equal to 3x cubed. To solvethis, subtract x cubed from both sides of the equation.To solve the equation x to the sixth minus 2x cubed is equal to 0,we can factor out an x cubed and obtain x cubed minus 2 equals 0.So we find that x equals 0 and y equals 0.Therefore, if x equals 0, then y equals 0. Since we already saidthat no solution exists for this equation when y = 0, then weshould ignore this value of x. We can then say that x cubed minus2 equals 0. Therefore, x = the cube root of 2.