Tangent lines to implicitly defined curves

The point belongs to the curve, since $F(1,1)=0$. The gradient at the point is $\nabla F(1,1)=(3,3)$, which is different from the zero vector, therefore the tangent line exists and has equation $F_x(1,1)(x-1)+F_y(1,1)(y-1)=0$. Dividing by $3$ we obtain $x+y=2$, that is $y=-x+2$. The same result is reached with Dini’s theorem: since $F_y(1,1)=3\ne 0$, near the point the curve is the graph of an implicit function with ${\phi }^{\prime }(1)=-\frac{3}{3}=-1$, and the tangent is $y=1-(x-1)$.

Substituting the coordinates of the point we obtain $F(1,1)=0$, hence the point belongs to the curve. In computing $F_x$ the term $x^2{y}^3$ is differentiated with respect to $x$ holding $y$ constant, and gives $2x{y}^3$; the term $x^3$ gives $3x^2$; at the point the sum equals $5$. In computing $F_y$ the same term gives $3x^2{y}^2$ and the term $-2y$ gives $-2$, with value $1$ at the point. Since the gradient $(5,1)$ is not zero, the tangent line has equation $5(x-1)+(y-1)=0$; solving for $y$ we obtain $y=-5x+6$. The slope $-5$ is the value ${\phi }^{\prime }(1)=-\frac{F_x}{F_y}$ of the derivative of the implicit function.

The point belongs to the curve because $e^0=1$ and $1-0-1=0$. The partial derivatives at the point are $F_x(0,1)=0$ and $F_y(0,1)=-1$: the gradient $(0,-1)$ is not zero, therefore the tangent exists, and since $F_y(0,1)\ne 0$ Dini’s theorem applies with respect to $y$. The derivative of the implicit function is ${\phi }^{\prime }(0)=-\frac{0}{-1}=0$, therefore the tangent line is horizontal and has equation $y=1$. The vanishing of $F_x$ at the point means that $x=0$ is a stationary point of the implicit function; the nature of such points is decided with the second derivative.