x^2 - 5xy + 3y^2 = 7

We need to differentiate the equation with respect to x.

==> dy/dx( x^2 - 5xy + 3y^2 ) = dy/dx ( 7)

==> dy/dx (x^2) - 5 dy/dx ( xy) + dy/dx ( 3y^2) = dy/dx ( 7).

We will differentiate with respect to x.

==> 2x - 5 ( 1*y + xy' ) + 6yy' = 0

==> 2x - 5y - 5xy'+ 6yy' = 0

Now we will combine the terms with y' on the left side.

==>  6yy' - 5xy' = 5y - 2x

Now we will factor y'.

==> y'*( 6y - 5x ) = 5y-2x

Now we will divide by (6y-5x).

==> y' = (5y-2x)/(6y-5x)

We are given the equation x^2-5xy+3y^2=7 and we have to find dy/dx.

Differentiating each term of x^2 - 5xy + 3y^2 = 7 with respect to x, we get

2x - 5x(dy/dx)- 5y + 6y(dy/dx) = 0

taking the terms with dy/dx to one side and the other terms to the opposite side we get

=> 2x - 5y = (dy/dx)( 5x - 6y)

divide both the sides by 5x - 6y

=> (dy/dx) = (2x - 5y) / ( 5x - 6y)

Therefore (dy/dx) = (2x - 5y) / ( 5x - 6y)