Equation of the given curve,
`x^(2/3)+y^(2/3) = 2`
Differentiating it w.r.t. x,
`2/3x^(-1/3)+2/3y^(-1/3)dy/dx = 0`
`=>dy/dx = -(x/y)^(-1/3) = -(y/x)^(1/3)`
`:. ` Slope of the tangent at point `(1,1) = m_1 = dy/dx = -(1)^(1/3) = -1`
Let slope of the normal is `m_2`. then,
`m_1*m_2 = -1 => -1*m_2 = -1 => m_2 = 1`
Now, equation of the tangent,
`y-1 = -1(x-1) => x+y-2 = 0`
Equation of the normal,
`y-1 = x-1 => x-y = 0.`