For each curve, plug in the given point [tex](x,y)[/tex] and check if the equality holds. For example:
(I) (2, 3) does lie on [tex]x^2+xy-y^2=1[/tex] since 2^2 + 2*3 - 3^2 = 4 + 6 - 9 = 1.
For part (a), compute the derivative [tex]\frac{\mathrm dy}{\mathrm dx}[/tex], and evaluate it for the given point [tex](x,y)[/tex]. This is the slope of the tangent line at the point. For example:
(I) The derivative is
[tex]x^2+xy-y^2=1\overset{\frac{\mathrm d}{\mathrm dx}}{\implies}2x+x\dfrac{\mathrm dy}{\mathrm dx}+y-2y\dfrac{\mathrm dy}{\mathrm dx}=0\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2x+y}{2y-x}[/tex]
so the slope of the tangent at (2, 3) is
[tex]\dfrac{\mathrm dy}{\mathrm dx}(2,3)=\dfrac74[/tex]
and its equation is then
[tex]y-3=\dfrac74(x-2)\implies y=\dfrac74x-\dfrac12[/tex]
For part (b), recall that normal lines are perpendicular to tangent lines, so their slopes are negative reciprocals of the slopes of the tangents, [tex]-\frac1{\frac{\mathrm dy}{\mathrm dx}}[/tex]. For example:
(I) The tangent has slope 7/4, so the normal has slope -4/7. Then the normal line has equation
[tex]y-3=-\dfrac47(x-2)\implies y=-\dfrac47x+\dfrac{29}7[/tex]
