not quite. You have some pieces but you’re putting them together incorrectly and assuming things you shouldn’t.

We don’t know that f(x,y) is a line , it doesn’t have to be. y=slope x + b, the tangent line is a separate equation from the function f.

Let’s call the tangent line z just to avoid reusing variables.

So z is a line of the form z= mx + C, where C is some constant. The slope of the tangent line is given by the derivative of f. And it hits the function at the point (X,Y) as stated in the question (I’m using capital X,Y to denote a point, lowercase to denote variable).

So the slope of the tangent line is exactly f’(X,Y)

Then we have the unknown constant in the line equation. You are told that the point (a,b) lies on the line so to find the value of the constant you plug in a point on the line:

z = m + C — slope m is the derivative of f at (X,Y)

z = f’(X,Y) x + C —plug in point (a,b)

b = f’(X,Y) a + C

C = b - f’(X,Y) a

So the final equation of the tangent is:

z = f’(X,Y) x + (b - f’(X,Y) a )